Solving the Difficult Integral ##\int_0^{\infty} x^{n+1} e^{-x} \sin(ax) dx##

[ergospherical]

Anyone have some ideas to approach the integral $\int_0^{\infty} x^{n+1} e^{-x} \sin(ax) dx$?

 

[fresh_42]

Well, already $a=1$ looks a bit unpleasant:

https://www.wolframalpha.com/input?i=integral+(from+0+to+infinity)+x^(n+1)+e^(-x)+sin(x)+dx=

Maybe you can find ideas in that series (there are several threads about integration)
https://www.physicsforums.com/threads/micromass-big-integral-challenge.867904/

 

[pasmith]

Anyone have some ideas to approach the integral $\int_0^{\infty} x^{n+1} e^{-x} \sin(ax) dx$?

Using $$\sin ax = \frac1{2i}(e^{iax} - e^{-iax})$$ we express the integral as a sum of integrals of the form $$I_n(c) = \int_0^\infty x^n e^{cx}\,dx$$ for complex $c$ with $\operatorname{Re}(c) < 0$. Then integrating by parts for $n > 0$ we obtain $$\begin{split}<br /> I_n(c) &= \left[\frac 1c x^ne^{cx}\right]_0^\infty - \frac{n}{c}I_{n-1}(c) \\<br /> &= -\frac nc I_{n-1}(c)<br /> \end{split}$$ and thus $$I_n(c) = (-1)^n\frac{n!}{c^n}I_0(c).$$ Then $$\begin{split}<br /> \int_)^\infty x^{n+1}e^{-x} \sin ax\,dx &= <br /> \frac {I_{n+1}(-1+ai) - I_{n+1}(-1-ai)}{2i} \\<br /> &= \frac{(-1)^{n+1}(n+1)!}{2i}\left(\frac{I_0(-1+ai)}{(-1+ai)^{n+1}} - \frac{I_0(-1-ai)}{(-1-ai)^{n+1}}\right).\end{split}$$

 

[topsquark]

To add to pasmith's idea:

Or, slightly more simply, use $sin(ax) = Im[ e^{iax}]$.

Then
$\displaystyle \int_0^{\infty} x^{n+1} e^{-x} \, sin(ax) \, dx = Im \left [ \int_0^{\infty} x^{n+1} e^{-x + iax} \, dx \right ]$

-Dan

 

[renormalize]

Or, slightly more simply, use $sin(ax) = Im[ e^{ia}]$.

Or perhaps $sin(ax) = Im[ e^{iax}]$?

 

[topsquark]

Or perhaps $sin(ax) = Im[ e^{iax}]$?

Thanks for the catch!

-Dan