Hi Kya Delhi 42 ki sabhi calculation plank level par mil sakti hain Gravity Acceleration Area Mass Density Energy density Angular velocity Angular energy Mean deviation from axis Potential energy Kinetic energy Tolal solid angle ratio with rest of the world Total mass contains Total elasticity Hooks law Total stress and strain Total g values according to solid angular velocity So we found how much energy we feel during earthquake if we are sitting at third floor Building is made with concrete Thanks No money but ideas are valuable For each particles of nation
I came upon this while studying variational approximation. Spent an entire hour integrating it with all the basic calculus I knew. Still got nothing. Had a mental breakdown and then fetched up my old friend from undergrad class at 12am. Ended up giving him brain attacks. Poor man is dying. I have to give him the link to this video. You save lives Andrew. Thank you so much❤️🎉. Edit: Yes. More math videos please!!!🙏.
Thank you for posting this, and going through all the algebra so slowly, and step by step. It's the best explanation I've seen so far, I finally get it now.
I myself am in the last year of my high school and I love physics and maths outside the school curriculum and this is probably the best it can get! Love your math and physics videos and keep making more good content Andrew 👍🏻👍🏻
hey, I currently am in the situation that you were in four years ago. I want to study engineering or physics and just couldn't help but wonder what you're doing now, four years after having probably the exact feeling that i have now
Dude you're so calm in your vídeos it makes me calm, while also learning stuff I find very nice and interesting but don't even need (I study chemical engineering). Anyway, I just discovered this channel and love it. Keep up the amazing work!
Would it not have been easier to use a gamma function here? To simplify the exponential power issue? It’s a nice channel you have here though Andrew and I love seeing a physics major discussing rigorous math methods. Keep it up!
I'm in high school and got to know about this Gaussian integral randomly. I surfed through the Internet for the derivation or evaluation but couldn't understand. But After watching your video it's all clear. Thanks & Keep Uploading.
Awww I was looking forward to you integrating the complex probability distribution Gaussian from the Uncertainty Principle's defined momentum spectrum to show us the quantum interference pattern of a particle or a set of particles using their mean value:)
Why is the angle for the integral from 0 to infinity equal to Pi/2? The integral evaluated with R covers the first quadrant (pi/2) but taking the analogy that it has ti rotate then that would be Pi. Even more so, this is half the first integral which had angle 2Pi, so for me it makes more sense for it to be Pi. Though I know it should be Pi/2, I just really don’t understand why
You mean around 9:40? It seems to me that you know that the angle needed to go from the x axis to the y axis is pi/2, but you also think that this integral should cover "half" of the plane that the first integral did, since the bounds are from 0 to infinity instead of from negative infinity to positive infinity. Am I understanding you? If I am, I think the problem you are having is that you are thinking one dimensionally instead of two. If only y was changed to be from 0 to infinity, the plane would be split in half so that only quadrants 1 and 2 are included and the bounds for theta would be from 0 to pi. However, BOTH the y bounds AND the x bounds are getting cut "in half" to be from 0 to infinity. This forces the integral to be over the first quadrant, or 1/2•1/2=1/4 the plane.
The little explanation at the beginning that justifies using y just tied this all together. I was wondering why you could just dump a different variable in and call it a day.
It would be nice to get a bit of a discussion about how strange it is that the areas and volumes of these functions are connected to pi when they seem to have nothing to do with pi, on the surface. Is it not noteworthy that pi magically appears, seemingly out of nowhere? It would also be nice to see some mention of how unusual it is that a cross-section of the solid you found the volume of, has an area of exactly the square root of the volume. What other solid has this property? These functions are very unusual, especially considering the connection to real world probability distributions. I don't mean to just complain, so I'll end on a positive note. Nice video!
What exactly is int_exp(-y^2) ? How can it be the same as int_exp(-x^2). I'm having a hard time interpreting what subbing y does. Do they both have the same graph?. If we can convert it to polar then doesn't y have to refer to the vertical axis? In that case how can the two functions be the same?
If you havnt already, look up metric coefficients. For it to actually be something, say a distance, you need d phi to be multiplied by some distance, in this case r. Thats how I intuitively remember it, but the wiki article will explain it alot better
This has to do with multivariable calculus. Reminder how in single variable calc how dx represented a small change in length, and we multiplied the function at the point times dx (infinitely) many times to find the area under the curve? We are integrating e^-(x^2+y^2), a function that depends on two variables. Since Andrew is doing a double integral, instead of integrating over length, he is integrating over a small patch of area dA in the plane. dA can be written as dxdy. Now we want to find a way to see how to write dA in terms of r and theta. This isn't completely rigorous, but it's how I see it: r and theta point at a single point in the plane, and r is the radius of a circle centered at (0,0). If we increase theta by dtheta, then the point moves a distance r•dtheta along the circle of radius r. Assuming that r is big, we can pretend that that length on the circle is a straight line. This is going to be the base of a rectangle of area dA. Now imagine increasing r by dr. This would give the rectangle a height dr. Therefore the area of the rectangle is dA= (r•dtheta)•dr = r•dr•dtheta, and since dA=dxdy, then dxdy=r•dr•dtheta.
Does anyone know a justification for turning the product of the two integrals into a double integral? I was confused and now am curious whether the following reasoning is valid: Want to find integral of f(x) from a to b, times integral g(y) from c to d. This is equal to { F(b) - F(a) } • { G(d) - G(c) }. Then I found the double integral of f(x)•g(y), assuming that x and y have no dependence on each other so that f(x) can be treated as a constant in y, and g(y) can be treated as a constant in x. Integrating with respect to x first I found integral with respect to y from c to d of g(y)•(F(b) - F(a)) Integrating once more, I found the same expression (F(b) - F(a)) • (G(d) - G(c)). Am I missing something? I haven't practiced my double integrals for a while.
This has to do with multivariable calculus. Reminder how in single variable calc how dx represented a small change in length, and we multiplied the function at the point times dx (infinitely) many times to find the area under the curve? We are integrating e^-(x^2+y^2), a function that depends on two variables. Since Andrew is doing a double integral, instead of integrating over length, he is integrating over a small patch of area dA in the plane. dA can be written as dxdy. Now we want to find a way to see how to write dA in terms of r and theta. This isn't completely rigorous, but it's how I see it: r and theta point at a single point in the plane, and r is the radius of a circle centered at (0,0). If we increase theta by dtheta, then the point moves a distance r•dtheta along the circle of radius r. Assuming that r is big, we can pretend that that length on the circle is a straight line. This is going to be the base of a rectangle of area dA. Now imagine increasing r by dr. This would give the rectangle a height dr. Therefore the area of the rectangle is dA= (r•dtheta)•dr = r•dr•dtheta, and since dA=dxdy, then dxdy=r•dr•dtheta.
This has to do with multivariable calculus. Reminder how in single variable calc how dx represented a small change in length, and we multiplied the function at the point times dx (infinitely) many times to find the area under the curve? We are integrating e^-(x^2+y^2), a function that depends on two variables. Since Andrew is doing a double integral, instead of integrating over length, he is integrating over a small patch of area dA in the plane. dA can be written as dxdy. Now we want to find a way to see how to write dA in terms of r and theta. This isn't completely rigorous, but it's how I see it: r and theta point at a single point in the plane, and r is the radius of a circle centered at (0,0). If we increase theta by dtheta, then the point moves a distance r•dtheta along the circle of radius r. Assuming that r is big, we can pretend that that length on the circle is a straight line. This is going to be the base of a rectangle of area dA. Now imagine increasing r by dr. This would give the rectangle a height dr. Therefore the area of the rectangle is dA= (r•dtheta)•dr = r•dr•dtheta, and since dA=dxdy, then dxdy=r•dr•dtheta.
This has to do with multivariable calculus. Reminder how in single variable calc how dx represented a small change in length, and we multiplied the function at the point times dx (infinitely) many times to find the area under the curve? We are integrating e^-(x^2+y^2), a function that depends on two variables. Since Andrew is doing a double integral, instead of integrating over length, he is integrating over a small patch of area dA in the plane. dA can be written as dxdy. Now we want to find a way to see how to write dA in terms of r and theta. This isn't completely rigorous, but it's how I see it: r and theta point at a single point in the plane, and r is the radius of a circle centered at (0,0). If we increase theta by dtheta, then the point moves a distance r•dtheta along the circle of radius r. Assuming that r is big, we can pretend that that length on the circle is a straight line. This is going to be the base of a rectangle of area dA. Now imagine increasing r by dr. This would give the rectangle a height dr. Therefore the area of the rectangle is dA= (r•dtheta)•dr = r•dr•dtheta, and since dA=dxdy, then dxdy=r•dr•dtheta. P.S. How'd you write theta?
@@Fysiker For theta you can open "character map" from your search bar and you'll find all the Greek letters, among other things. I actually did some research into this awhile ago and understand it to come from finding the Jacobian where x = r cosθ and y = r sin θ. In this case, the Jacobian being the determinant of the 2x2 matrix: [ ∂x/∂r ∂x/∂θ ] [ ∂y/∂r ∂y/∂θ ] Thank you for your help
I didn't understand how you can let x' equal (x^2)/b? Why isn't it -2x/b? These videos are very engaging and useful by the way. Thank you Andrew Dotson!
I'm a little late probably. I am assuming you had x' as a notation for the derivative. In the video x' is not the derivative of x, it's just the name of the variable.
@@AndrewDotsonvideos I suppose it does! But before you do the integral is there a way to tell whether it will actually converge or you just have to go through the process to find out (e.g. for instance could you draw the graph & sort of "guess" from looking at it that the integral likely exists?)
Why did you square the integral first? Every proof I've seen uses this trick in one guise or another but no one has explained how you would know to do that before knowing the answer you're looking for
You have to square the integral in order to be able to transform the exponent into polar coordinates. It's also not exactly something you're supposed to know to do. People had been trying to solve this unsuccessfully for almost 70 years before rootin' tootin' Gauss did the polar coordinates transform.
Well you can do it much more simply man.e^x2 is an even function.so limit of integration becomes 0 to inf with a multiplication of 2 infront of the integral. Then use x^2 =t. Then it is simply a gamma function,which can be evaluated without even integrating.Lol
Your integral signs are masterpieces
I just noticed a mistake at 11:16. dx' does not equal dx. dx' = dx/ root b! Sorry about that. So it should be multipled by root b times dx'.
Hi
Kya Delhi 42 ki sabhi calculation plank level par mil sakti hain
Gravity
Acceleration
Area
Mass
Density
Energy density
Angular velocity
Angular energy
Mean deviation from axis
Potential energy
Kinetic energy
Tolal solid angle ratio with rest of the world
Total mass contains
Total elasticity Hooks law
Total stress and strain
Total g values according to solid angular velocity
So we found how much energy we feel during earthquake if we are sitting at third floor
Building is made with concrete
Thanks
No money but ideas are valuable
For each particles of nation
Fun fact, the Fourier transform of a Gaussian is another Gaussian.
Kinda expected, but fun !
I came upon this while studying variational approximation. Spent an entire hour integrating it with all the basic calculus I knew. Still got nothing. Had a mental breakdown and then fetched up my old friend from undergrad class at 12am. Ended up giving him brain attacks. Poor man is dying. I have to give him the link to this video. You save lives Andrew. Thank you so much❤️🎉.
Edit: Yes. More math videos please!!!🙏.
When I learn this in school I get bored
But when I find this on RUclips I find it really interesting
how strange
Thank you for posting this, and going through all the algebra so slowly, and step by step. It's the best explanation I've seen so far, I finally get it now.
Probably one of my favorite videos so far.
I myself am in the last year of my high school and I love physics and maths outside the school curriculum and this is probably the best it can get! Love your math and physics videos and keep making more good content Andrew 👍🏻👍🏻
hey, I currently am in the situation that you were in four years ago. I want to study engineering or physics and just couldn't help but wonder what you're doing now, four years after having probably the exact feeling that i have now
More math walkthroughs!
Dude you're so calm in your vídeos it makes me calm, while also learning stuff I find very nice and interesting but don't even need (I study chemical engineering). Anyway, I just discovered this channel and love it. Keep up the amazing work!
Thanks for the nice comment!
"Everyone's favourite integral to look up" so true lol 😂
Would it not have been easier to use a gamma function here? To simplify the exponential power issue?
It’s a nice channel you have here though Andrew and I love seeing a physics major discussing rigorous math methods. Keep it up!
I'm in high school and got to know about this Gaussian integral randomly. I surfed through the Internet for the derivation or evaluation but couldn't understand. But After watching your video it's all clear.
Thanks & Keep Uploading.
Really fun! Thanks :) I hope you do more similar math related videos.
loved it thank you for taking time out and making these vids. would love to see more math videos
I send thanks from Griffiths Intro to Quantum Mechanics 3e problem 1.3
Just discovered Papa Flammy the other week and now I can get my fix from this channel too!
Welcome!
Amazing video! Hoping to see more math related videos in the (near) future :)
I love how you spun around to represent 2pi.
The general way of going about solving this is to be A LItTle CleVEr
Awww I was looking forward to you integrating the complex probability distribution Gaussian from the Uncertainty Principle's defined momentum spectrum to show us the quantum interference pattern of a particle or a set of particles using their mean value:)
So I'm in precalculus 11 right now but this was still fun to watch hahaha cool vid man!
Awesome....it's good to have someone who explains stuff you don't wanna read from the book...lol.
Awesome!! Thanks to you.I was struggling with this for so long.
Very good explanation! I love this channel!
A slick way to normalize the Gaussian distribution. Error functions have no closed form.
One little mistake r = x^2 + y^2 should be r^2 instead of r. You made the correct substitution into the integral though.
Why is the angle for the integral from 0 to infinity equal to Pi/2? The integral evaluated with R covers the first quadrant (pi/2) but taking the analogy that it has ti rotate then that would be Pi. Even more so, this is half the first integral which had angle 2Pi, so for me it makes more sense for it to be Pi. Though I know it should be Pi/2, I just really don’t understand why
You mean around 9:40? It seems to me that you know that the angle needed to go from the x axis to the y axis is pi/2, but you also think that this integral should cover "half" of the plane that the first integral did, since the bounds are from 0 to infinity instead of from negative infinity to positive infinity. Am I understanding you?
If I am, I think the problem you are having is that you are thinking one dimensionally instead of two. If only y was changed to be from 0 to infinity, the plane would be split in half so that only quadrants 1 and 2 are included and the bounds for theta would be from 0 to pi. However, BOTH the y bounds AND the x bounds are getting cut "in half" to be from 0 to infinity. This forces the integral to be over the first quadrant, or 1/2•1/2=1/4 the plane.
You cleaned the whiteboard
Great work,it was so joyous to learn this incredible beauty👍
is there a Fubinni hidden somewhere ?
I tried to solve it using euler's identity but got stuck for the infinity limits. This method helps and was wonderful to learn. Thank you.
You’re an excellent teacher
Why dxdy=rdrdθ?
PI apparences in EXp integral always still amaze me
The little explanation at the beginning that justifies using y just tied this all together. I was wondering why you could just dump a different variable in and call it a day.
I have an entrance exam on Wednesday, this was a good refresher
Wow..very nice explanation..I would love to see more videos like this ;)
You'll be a good prof fr 🤔
make more math videos! I love the way you explain!
when you replace dxdy wheres the r come from?why is the replacement to dxdy, r * drdtheta?
It would be nice to get a bit of a discussion about how strange it is that the areas and volumes of these functions are connected to pi when they seem to have nothing to do with pi, on the surface. Is it not noteworthy that pi magically appears, seemingly out of nowhere? It would also be nice to see some mention of how unusual it is that a cross-section of the solid you found the volume of, has an area of exactly the square root of the volume. What other solid has this property? These functions are very unusual, especially considering the connection to real world probability distributions. I don't mean to just complain, so I'll end on a positive note. Nice video!
Ah that was neat! We were just given some Gaussian integrals in class and use those
Hey, I liked it, but I’m stuck where there’s an x in front of the Gaussian
what book do you suggest to learn new integrating method,amazing video.
You can just search for "how to do any integral" here on RUclips
What exactly is int_exp(-y^2) ? How can it be the same as int_exp(-x^2). I'm having a hard time interpreting what subbing y does. Do they both have the same graph?. If we can convert it to polar then doesn't y have to refer to the vertical axis? In that case how can the two functions be the same?
1:58 Missed the dx
Deleting my channel right now
Andrew Dotson good. Jk. Loved the video. This was entertaining. And you caught your minor mistakes. Nice job man.
Really appreciate it!
Could you explain why dxdy became rdrdtheta?
If you havnt already, look up metric coefficients. For it to actually be something, say a distance, you need d phi to be multiplied by some distance, in this case r. Thats how I intuitively remember it, but the wiki article will explain it alot better
Polar coordinates.
Look up the jacobian, you can think of it as a compensation factor for the transformation between coordinate systems
When you take calc 3 you will see why, in brief, it is just a substitution for dxdy or dydx.
This has to do with multivariable calculus. Reminder how in single variable calc how dx represented a small change in length, and we multiplied the function at the point times dx (infinitely) many times to find the area under the curve? We are integrating e^-(x^2+y^2), a function that depends on two variables.
Since Andrew is doing a double integral, instead of integrating over length, he is integrating over a small patch of area dA in the plane. dA can be written as dxdy. Now we want to find a way to see how to write dA in terms of r and theta.
This isn't completely rigorous, but it's how I see it: r and theta point at a single point in the plane, and r is the radius of a circle centered at (0,0). If we increase theta by dtheta, then the point moves a distance r•dtheta along the circle of radius r. Assuming that r is big, we can pretend that that length on the circle is a straight line. This is going to be the base of a rectangle of area dA.
Now imagine increasing r by dr. This would give the rectangle a height dr. Therefore the area of the rectangle is dA= (r•dtheta)•dr = r•dr•dtheta, and since dA=dxdy, then
dxdy=r•dr•dtheta.
Would you please cover Feynman’s method?
Me wondering how he draws perfect integrals, every time
why do we use the delta-distribution as a sequence of gaussian functions instead of just defining a function that returns 0 and 1 as in maths?
Does anyone know a justification for turning the product of the two integrals into a double integral? I was confused and now am curious whether the following reasoning is valid:
Want to find integral of f(x) from a to b, times integral g(y) from c to d. This is equal to
{ F(b) - F(a) } • { G(d) - G(c) }.
Then I found the double integral of f(x)•g(y), assuming that x and y have no dependence on each other so that f(x) can be treated as a constant in y, and g(y) can be treated as a constant in x. Integrating with respect to x first I found
integral with respect to y from c to d of g(y)•(F(b) - F(a))
Integrating once more, I found the same expression (F(b) - F(a)) • (G(d) - G(c)).
Am I missing something? I haven't practiced my double integrals for a while.
Glorious! Woulda taken me a couple minutes to think to swap to polar coordinates!
I the end it should be \sqrt{b} x prime
Right?
Using the gamma function is my favourite though
Why did he change the dx into r•dr?
This has to do with multivariable calculus. Reminder how in single variable calc how dx represented a small change in length, and we multiplied the function at the point times dx (infinitely) many times to find the area under the curve? We are integrating e^-(x^2+y^2), a function that depends on two variables.
Since Andrew is doing a double integral, instead of integrating over length, he is integrating over a small patch of area dA in the plane. dA can be written as dxdy. Now we want to find a way to see how to write dA in terms of r and theta.
This isn't completely rigorous, but it's how I see it: r and theta point at a single point in the plane, and r is the radius of a circle centered at (0,0). If we increase theta by dtheta, then the point moves a distance r•dtheta along the circle of radius r. Assuming that r is big, we can pretend that that length on the circle is a straight line. This is going to be the base of a rectangle of area dA.
Now imagine increasing r by dr. This would give the rectangle a height dr. Therefore the area of the rectangle is dA= (r•dtheta)•dr = r•dr•dtheta, and since dA=dxdy, then
dxdy=r•dr•dtheta.
@@Fysiker thanks for explaining this! This helped a lot.
Yay a math video I can understand!
Please make some videos on statistical mechanics.♡
How does dxdy = rdrdtheta ?
This has to do with multivariable calculus. Reminder how in single variable calc how dx represented a small change in length, and we multiplied the function at the point times dx (infinitely) many times to find the area under the curve? We are integrating e^-(x^2+y^2), a function that depends on two variables.
Since Andrew is doing a double integral, instead of integrating over length, he is integrating over a small patch of area dA in the plane. dA can be written as dxdy. Now we want to find a way to see how to write dA in terms of r and theta.
This isn't completely rigorous, but it's how I see it: r and theta point at a single point in the plane, and r is the radius of a circle centered at (0,0). If we increase theta by dtheta, then the point moves a distance r•dtheta along the circle of radius r. Assuming that r is big, we can pretend that that length on the circle is a straight line. This is going to be the base of a rectangle of area dA.
Now imagine increasing r by dr. This would give the rectangle a height dr. Therefore the area of the rectangle is dA= (r•dtheta)•dr = r•dr•dtheta, and since dA=dxdy, then
dxdy=r•dr•dtheta.
Take the determinant of the Jacobian matrix. This is used when changing variables
How u change r=x^2+y^2 to r^2?
You explained really well bro
This is the way that physicist do integration
PKMKB
I know I'm missing something minor here, but I'm not clear on why dxdy becomes rdrdθ instead of drdθ. Where did the r come from?
It comes from the jacobins:) dxdy has units of area but drdtheta only has units of length since angles are dimensionless.
@@AndrewDotsonvideos Oooh. I see. Thanks Andrew :)
Really enjoyed this need more
what is the integration of e^(2x-x^2)
Is there a general rule that the product of two integrals equals the double integral? Referring to 2:50.
fubini's theorem justify this i believe as long as both integrands are continuous within the region of integration
6:28 If we had a bottom limit, what would we multiply with?
(2pi-bottom limit), you won't have a bottom limit though.
Why does dxdy = r drdθ ?
This has to do with multivariable calculus. Reminder how in single variable calc how dx represented a small change in length, and we multiplied the function at the point times dx (infinitely) many times to find the area under the curve? We are integrating e^-(x^2+y^2), a function that depends on two variables.
Since Andrew is doing a double integral, instead of integrating over length, he is integrating over a small patch of area dA in the plane. dA can be written as dxdy. Now we want to find a way to see how to write dA in terms of r and theta.
This isn't completely rigorous, but it's how I see it: r and theta point at a single point in the plane, and r is the radius of a circle centered at (0,0). If we increase theta by dtheta, then the point moves a distance r•dtheta along the circle of radius r. Assuming that r is big, we can pretend that that length on the circle is a straight line. This is going to be the base of a rectangle of area dA.
Now imagine increasing r by dr. This would give the rectangle a height dr. Therefore the area of the rectangle is dA= (r•dtheta)•dr = r•dr•dtheta, and since dA=dxdy, then
dxdy=r•dr•dtheta.
P.S. How'd you write theta?
@@Fysiker For theta you can open "character map" from your search bar and you'll find all the Greek letters, among other things.
I actually did some research into this awhile ago and understand it to come from finding the Jacobian where x = r cosθ and y = r sin θ.
In this case, the Jacobian being the determinant of the 2x2 matrix:
[ ∂x/∂r ∂x/∂θ ]
[ ∂y/∂r ∂y/∂θ ]
Thank you for your help
i can smell the whiteboard marker
I didn't understand how you can let x' equal (x^2)/b?
Why isn't it -2x/b?
These videos are very engaging and useful by the way. Thank you Andrew Dotson!
I'm a little late probably. I am assuming you had x' as a notation for the derivative. In the video x' is not the derivative of x, it's just the name of the variable.
I think first show that the integral exists (i.e. converges) & why then get into the mechanics of integration.
Robert Munga does solving the integral not show that it converges?
@@AndrewDotsonvideos I suppose it does! But before you do the integral is there a way to tell whether it will actually converge or you just have to go through the process to find out (e.g. for instance could you draw the graph & sort of "guess" from looking at it that the integral likely exists?)
Thank you teacher
Excellent explanation!
I like the math videos but I also think you should definitely make more joke videos
YOooo! where are the pictures...son?
could 'e' be any constant? if not why not?
Because e is a constant, not a variable...
thank you thank you thank you thank youuuuuuuu
isn't it easier to use the power series for e^x and integrate the series?
I haven't tried that! Give it a try and let me know how it goes!
So when is the exam?
Helped a lot, thank you!!!
very nice! 👍👏👏👏
Thank you 👍👍
...or just use a symmetry argument to say that the integral from 0 to infinity is exactly half the integral from -infinity to infinity.
Since exp(-x^2) is even***, i.e., exp(-(-x)^2 ) = exp(-(x)^2)
Great video, thanks 👍🏼
What is the meaning of that symbol you call “sine”? i dont know how to spell it either
Pham Huu Tri Its pronounced as Psy(sy).
Thanks Andrew
I'm Just learning differentiation in highschool math. This looks terrifying.
More than terrifying it is terrific
Great video!
You are a genius.....
This would’ve been useful a day ago for my quantum final. Rip.
so helpful thanks
Why did you square the integral first? Every proof I've seen uses this trick in one guise or another but no one has explained how you would know to do that before knowing the answer you're looking for
You have to square the integral in order to be able to transform the exponent into polar coordinates.
It's also not exactly something you're supposed to know to do. People had been trying to solve this unsuccessfully for almost 70 years before rootin' tootin' Gauss did the polar coordinates transform.
Its easy, just look at the table of integrals
Well you can do it much more simply man.e^x2 is an even function.so limit of integration becomes 0 to inf with a multiplication of 2 infront of the integral. Then use x^2 =t. Then it is simply a gamma function,which can be evaluated without even integrating.Lol
? The gamma function has another factor of t^(z-1) though.
❤️❤️❤️❤️❤️
Thank you!
Why are math guys always cute?
more of this kind of video
Ta da ... ❤
Thanks!