A bit "wrong" way to do it is remeber that sin(x) = im(exp(ix)) and do a change of variables: ix = -at. With this, only remains calculate the Laplace transform and take the imaginary part.
I used tablature integration to derive the finite series representation of the Mellin transform. I then evaluated the series at k = s-1 because the other terms 0 out. The answer becomes sin(pi*k/2)*Gamma(s)/Gamma(s-k+1) evaluated at k = s so Gamma(s)sin(pi*k/2) is the final answer
3:16 I miss referencing gamelin! But I'll leave the reference as an exercise for the reader and instead just say awesome vid mah dude! I think we can also create a left hand inequality to say that the imaginary part of ((-i)^-s*Gamma(s)) = Gamma(s) sin(pis/2) of which we perform the adequate procedures or rigor so that we don't make Gamelin angry.
multiply the integrand by e/e, you then get 1/e int 0 to 1 e^(x+1) / x+1 which is simply the exponential integral evaluated from 1 to 2, so your answer is (Ei(2)-Ei(1))/e
Just some ideas... First I noticed that ln(x + sqrt(x^2+1)) is equal to arsinh(x). Hence substituting u = arsinh(x) transforms the integral into one from 0 to arsinh(1) over u times tanh(u) du. Then I tried integration by parts; now I'm left with an integral over ln(cosh(u)). I think solving that integral probably works similarly to the integral over ln(cos(x)), the latter one can be found in many places... Hope that helps. Probably there is an easier way, but that's the first ideas which came to my mind.
@@bjornfeuerbacher5514 I'm only in class 12th so I don't have the best knowledge about improper integral And I don't even know if this is doable, but I did see its graph and it seems to be convergent Thank you
@@bjornfeuerbacher5514 tanh(u) = (exp(2u)-1)/(exp(2u)+1).= exp(-2u) (exp(2u)-1)/(1+exp(-2u)). Then you could use the serie expansion of 1/(1+exp(-2u)) in power of exp(-2u), I guess ? You would be left with integrating a bunch of u*exp(-k u) and exp(-k u) factors I think.
@@shivanshnigam4015your integral can be shown to be equal to int 0 to arcsinh(1) of -ln(sinh(x)) which can be evaluated by using the exponential definition of sinh(x), separating the ln(2), factoring out an e^x term from ln(e^x-e^-x), then invoking the series expansion of the log, leaving you with an integral with an antideriative -(1/2)x^2+xln(2)-Li_2(e^-2x)/2, which when evaluated from 0 to arcsinh(1) gives you the answer (1/12)pi^2 - (1/2)arcsinh^2(1) + ln(2)arcsinh(1) -(1/2)Li_2(3-2sqrt(2))
5:56 didn't expect that 🤨
Rock hard dikc
sinx = imaginary part of exp(ix)
that's what i thought he'd do
A bit "wrong" way to do it is remeber that sin(x) = im(exp(ix)) and do a change of variables: ix = -at. With this, only remains calculate the Laplace transform and take the imaginary part.
Very interesting approach for this integral. Note that for the substitution you should put du instead of dt. Thanks alot.
I used tablature integration to derive the finite series representation of the Mellin transform. I then evaluated the series at k = s-1 because the other terms 0 out. The answer becomes sin(pi*k/2)*Gamma(s)/Gamma(s-k+1) evaluated at k = s so Gamma(s)sin(pi*k/2) is the final answer
3:16 I miss referencing gamelin! But I'll leave the reference as an exercise for the reader and instead just say awesome vid mah dude!
I think we can also create a left hand inequality to say that the imaginary part of ((-i)^-s*Gamma(s)) = Gamma(s) sin(pis/2) of which we perform the adequate procedures or rigor so that we don't make Gamelin angry.
couldn't It be done by im(e^ix) for sinx and a substitution to directly use the gamma function from the definition?
Hi,
Nice. What can we do with this transformation?
My favorite transform
Beautiful! And Fun!
Well done buddy!
Master can you solve my integration doubt please 🙏
Int 0 to 1 [ e^x / (x+1) ] dx
multiply the integrand by e/e, you then get 1/e int 0 to 1 e^(x+1) / x+1 which is simply the exponential integral evaluated from 1 to 2, so your answer is (Ei(2)-Ei(1))/e
Nice bro❤
When s=0, the result is incorrect. Gamma(0) is not defined.
The limit Gamma(s)*sin(pi*s) exists
Yes and using the famous sin(x)/x limit and the recursion formula for the gamma function, we get pi/2 as guaranteed by the Dirichlet integral.
The Merlin transform 🪄
The condition must be 0
No...-1
@@DihinAmarasigha-up5hf but gamma of s-1 existes for s-1 >0
hey, please help me on this one
int 0 to 1 (ln(x + sqrt(x^2+1))/x
Just some ideas... First I noticed that ln(x + sqrt(x^2+1)) is equal to arsinh(x). Hence substituting u = arsinh(x) transforms the integral into one from 0 to arsinh(1) over u times tanh(u) du. Then I tried integration by parts; now I'm left with an integral over ln(cosh(u)). I think solving that integral probably works similarly to the integral over ln(cos(x)), the latter one can be found in many places...
Hope that helps. Probably there is an easier way, but that's the first ideas which came to my mind.
I'll give it a shot bro
@@bjornfeuerbacher5514 I'm only in class 12th so I don't have the best knowledge about improper integral
And I don't even know if this is doable, but I did see its graph and it seems to be convergent
Thank you
@@bjornfeuerbacher5514 tanh(u) = (exp(2u)-1)/(exp(2u)+1).= exp(-2u) (exp(2u)-1)/(1+exp(-2u)). Then you could use the serie expansion of 1/(1+exp(-2u)) in power of exp(-2u), I guess ? You would be left with integrating a bunch of u*exp(-k u) and exp(-k u) factors I think.
@@shivanshnigam4015your integral can be shown to be equal to int 0 to arcsinh(1) of -ln(sinh(x)) which can be evaluated by using the exponential definition of sinh(x), separating the ln(2), factoring out an e^x term from ln(e^x-e^-x), then invoking the series expansion of the log, leaving you with an integral with an antideriative -(1/2)x^2+xln(2)-Li_2(e^-2x)/2, which when evaluated from 0 to arcsinh(1) gives you the answer (1/12)pi^2 - (1/2)arcsinh^2(1) + ln(2)arcsinh(1) -(1/2)Li_2(3-2sqrt(2))
Wow
Why are likes hidden?
I honestly have no idea and I've noticed this in the last video too
Oh, anyways great video keep up the good work
I think "Melin" is misspelled in the title :)
Thank you
Fixed it