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wow the mic quality is so good

Looks interesting, but I can't figure out what steps you're skipping when you claim an bounds of zero and two for the integral wrt t.

specific heat of potato is about 3.4 kJ/kg

Here’s to hoping you continue this series

Nice one

Very interesting

WOAH!!! 🤯🤯🤯🤯

I've used f(z) = ln(z + ib) / (z² + a²) , b>0 And then contour integration by using upper side rectangular contour with residue at z = ia , and then equating real parts and we got it.

Integration by parts twice and substitution x = tan(t) leads us to the integral -4\int_{0}^{\frac{\pi}{2}}\ln{(cos{(t)})}dt

I'm curious to know how we can the pole at z=i because it is of infinite order and the power series can't be used as well due to the 1/(z+i) being there...any ideas on how to calculate it...?

I=int[-♾️,♾️]((1+x^2)^-(n+1))dx x=tan(y) dx=sec^2(y)dy I=int[-pi/2,pi/2](sec^(-2n)(y))dy I=2•int[0,pi/2](sin^0(y)cos^2n(y))dy beta(u,v)=2•int[0,pi/2](sin^(2u-1)(y)cos^(2v-1)(y))dy I=beta(1/2,n+1/2) I=sqrt(pi)gamma(n+1/2)/gamma(n+1)

y=x^t x=y^(1/t) dx=1/t•y^(1/t-1)•dy I=1/t^2•int[0,1](ln(y)ln(1-y)/y)dy u=ln(y) dv=ln(1-y)/y du=dy/y v=-Li_2(y) I=1/t^2•(-ln(y)Li_2(y)|[0,1]+int[0,1](Li_2(y)/y)dy) I=Li_3(1)/t^2 Li_3(1)=sum[k=1,♾️](1^k/k^3)=zeta(3) I=zeta(3)/t^2

what about the arcs tho?

I didn't use complex analysis but simple integration techniques and I got the result Integral(-½ to +½) Γ(1+x) Γ(1-x) dx = (4/π) β(2) = 4G/π Here β is dirichlet beta function.

Please dont say lohopitals)

Choise of contours always seems completely arbitrary to me ://

bro forgot to edit out the first take

L'Hopital)

Take the derivative of both sides to give y’ = y’’ + y’’’ + … Then plug into the original equation to get y = 2y’ Giving the solution y = Ae^ (x/2)

cool method before i watch the video i think that you will use the box contour but i suprise by using the semicircle contour really wonderful 🥰🥰

really cool🥰🥰 i love this contour integration🤩 but sir can you explain why the second integral =- pi res(0) because i couldn't really understand. thank you !! and keep going! more contour integration😊

Another way you could do it: find a power series for ln(1-x^t)/x and then you pretty much all that's left in the integral is ln(x) ⋅ power function which is easily handled by integration by parts

try and get rid of the annoying background noise in future videos

Really)

That other dude seems pretty cool

Genius

Academic WEAPON

Normal way is so much better than complex way

Reflection formula. I am genuinely surprised i know it)))

i used a quite different method which actually turned out simpler: i used the identity arctan(x)+arctan(1/x)=pi/2, to rewrite the starting integral as pi/2-arctan(1/x) all over x (and all squared), expanding the square we get 3 integrals, 2 of which trivially go to zero (odd functions), and we're left with the integral of arctan(1/x)^2/x^2, and being even i multiplied by 2 and changed the bounds to from 0 to inf i then subbed 1/x=u, which easily simplifies, and with some further calculations you end up with twice the integral from 0 to inf of ln(x^2+1)/(x^2+1), where subbing tan(t)=x transforms the integral into 4 times the integral from 0 to pi/2 of -ln(cost), which is a known integral which evaluates to -pi/2ln2. multiplying with the -4 in front we get 2*pi*ln2

cool, looked hard but turned out simple, i wouldn't have thought of using the other pole

Complex analysising is my new favourite verb

Finally thanks

Could there be a parametic function that creates a square?

finally. thank you.

underrated channel

Cool video, is this how they define the fractional derivatives?

I want to cry. contouring is still a makeup technique to me.

crazy...

mmm yes I definitely follow the logic 🧠

You want to say that 1+3=3^2

I miss physics

Why did 4 people dislike what is there to not like about this video

Fume hoods are cool. Use them.

i hope that there is exhaust for that fumes that it produces

Big boom

Safety, check

Purple flame must be potassium based