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
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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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1. What is the purpose of solving this integral?

The purpose of solving this integral is to find the definite integral of a function that involves exponential and trigonometric terms. This type of integral commonly appears in physics and engineering problems, making it important for scientific research and applications.

2. What is the general approach for solving this integral?

The general approach for solving this integral is to use integration by parts, where one part of the integrand is differentiated and the other is integrated. This method allows us to simplify the integral and potentially solve it using other integration techniques such as substitution or partial fractions.

3. Can this integral be solved analytically?

Yes, this integral can be solved analytically using the aforementioned integration by parts method. However, the final result may involve special functions such as the gamma function or the error function, which may require numerical methods for evaluation.

4. Are there any special cases for this integral?

Yes, there are special cases for this integral depending on the values of n and a. For example, if n is an even integer and a is an odd multiple of π, the integral evaluates to 0. Additionally, if n is a negative integer and a is a positive real number, the integral diverges.

5. How can this integral be applied in real-world problems?

This integral can be applied in various real-world problems, particularly in physics and engineering. For example, it can be used to calculate the energy of a quantum mechanical system or to model the damping effect in oscillating systems. It can also be used in signal processing to analyze the frequency spectrum of a signal.

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