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

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• ergospherical
In summary, to approach the integral ##\int_0^{\infty} x^{n+1} e^{-x} \sin(ax) dx##, we can use the identity ##\sin(ax) = \frac1{2i}(e^{iax} - e^{-iax})## and express it 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, using integration by parts and simplifying, we can obtain a final expression for the integral. Alternatively, we can use the identity ##\sin(ax
ergospherical
Anyone have some ideas to approach the integral ##\int_0^{\infty} x^{n+1} e^{-x} \sin(ax) dx##?

ergospherical said:
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 $$\begin{split} I_n(c) &= \left[\frac 1c x^ne^{cx}\right]_0^\infty - \frac{n}{c}I_{n-1}(c) \\ &= -\frac nc I_{n-1}(c) \end{split}$$ and thus $$I_n(c) = (-1)^n\frac{n!}{c^n}I_0(c).$$ Then $$\begin{split} \int_)^\infty x^{n+1}e^{-x} \sin ax\,dx &= \frac {I_{n+1}(-1+ai) - I_{n+1}(-1-ai)}{2i} \\ &= \frac{(-1)^{n+1}(n+1)!}{2i}\left(\frac{I_0(-1+ai)}{(-1+ai)^{n+1}} - \frac{I_0(-1-ai)}{(-1-ai)^{n+1}}\right).\end{split}$$

ergospherical, topsquark and jedishrfu

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

Last edited:
topsquark said:
Or, slightly more simply, use ##sin(ax) = Im[ e^{ia}]##.
Or perhaps ##sin(ax) = Im[ e^{iax}]##?

topsquark
renormalize said:
Or perhaps ##sin(ax) = Im[ e^{iax}]##?
Thanks for the catch!

-Dan

renormalize and fresh_42

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