A Mathematical Proof for an Improper Integral

Question

I got the following solution for the integral $I$ from Wolfram and I have verified the solution numerically which seems to be correct! Does someone have an idea about the mathematical proof?

$I = \int_{0}^{\infty} \frac{e^{-ax}sin(bx)}{x} dx = \arctan(\frac{b}{a})$

in which a and b are some positive constants.

Thanks in advance.

Thanks for your answer but the problem is that I cannot calculate the integral resulting from integration with respect to $b$, i.e. $\int \frac{b \sin(bx) - a \cos(bx)}{a^2+b^2} db$. — Shahram Khazaie, Nov 28 '13 at 17:21
You applied the limits incorrectly. It should be much simpler than that. I think you forgot to set $x$ to zero. — Stephen Montgomery-Smith, Nov 28 '13 at 17:23
You are absolutely right. Thank you very much for your answer. It helped me a lot. — Shahram Khazaie, Nov 28 '13 at 19:52

score 2 · Accepted Answer · answered Nov 27 '13 at 18:27

2

Start with $\int_0^\infty e^{-ax} \cos(bx) \, dx$, which is easily worked out. Then integrate with respect to $b$. Use $b=0$ to establish the integration constant. Wave your hands to explain the validity of exchanging the two integrals (or cite Fubini's Theorem).

answered Nov 27 '13 at 18:27

Stephen Montgomery-Smith

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You beat me to it (I was going to provide more details …) But more to the point, the path to discovering this proof involves differentiating the original integral with respect to $a$ in order to get rid of the bothersome $x$ in the denominator. Then realize that it is better to do this argument backwards because Fubini is easier to justify than differentiating under the integral sign. – Harald Hanche-Olsen Nov 27 '13 at 18:32
Yes, I must admit that I did it be first differentiating. And I'm sure I had seen the trick used somewhere else, and was dragging up stuff from my memory. – Stephen Montgomery-Smith Nov 27 '13 at 18:37

MathArt · Answer 2 · 2024-03-15T09:50:06.647

It might be straightforward to use the technique of Laplace transform, particularly from Laplace transform of the product of two functions.

\begin{align} I&=\int_{0}^{\infty} \frac{e^{-sx}\sin(bx)}{x}\mathrm dx\\ &=\int_{0}^{\infty}\mathcal{L^{-1}}\left[\frac1x\right](\xi)\mathcal{L}\left[\sin(bx)\right](s+\xi)\mathrm d\xi\\ &=\int_{0}^{\infty}1(\xi)\frac{b}{(s+\xi)^2+b^2}\mathrm d\xi\\&=\frac{\pi}{2}-\arctan(\frac{s}{b})\\&=\arctan(\frac{b}{s}). \end{align}

score 0 · Answer 3 · edited Apr 13 '17 at 12:20

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You can follow the steps

i) Differentiate $I$ w.r.t. $a$.

ii) evaluate the integral $I_a$.

iii) integrate with respect with $a$ and note that $\lim_{a\to \infty} I =0. $

Here is a related problem.

edited Apr 13 '17 at 12:20

Community

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answered Nov 27 '13 at 19:00

Mhenni Benghorbal

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A Mathematical Proof for an Improper Integral

3 Answers3