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

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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
  • #1
ergospherical
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Anyone have some ideas to approach the integral ##\int_0^{\infty} x^{n+1} e^{-x} \sin(ax) dx##?
 
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  • #3
ergospherical said:
Anyone have some ideas to approach the integral ##\int_0^{\infty} x^{n+1} e^{-x} \sin(ax) dx##?

Using [tex]\sin ax = \frac1{2i}(e^{iax} - e^{-iax})[/tex] we express the integral as a sum of integrals of the form [tex]I_n(c) = \int_0^\infty x^n e^{cx}\,dx[/tex] for complex [itex]c[/itex] with [itex]\operatorname{Re}(c) < 0[/itex]. Then integrating by parts for [itex]n > 0[/itex] we obtain [tex]
\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}[/tex] and thus [tex]
I_n(c) = (-1)^n\frac{n!}{c^n}I_0(c).[/tex] Then [tex]
\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}[/tex]
 
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  • #4
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
 
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  • #5
topsquark said:
Or, slightly more simply, use ##sin(ax) = Im[ e^{ia}]##.
Or perhaps ##sin(ax) = Im[ e^{iax}]##?
 
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  • #6
renormalize said:
Or perhaps ##sin(ax) = Im[ e^{iax}]##?
Thanks for the catch!

-Dan
 
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