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I'm trying to evaluate the integral below for my research related to sound radiation. Assume $a$ is a positive constant.

$$\int_{-\infty}^\infty\frac{e^{-ik\sqrt{x^2+a^2}}}{\sqrt{x^2+a^2}}dx$$

First, I tried to separate the functon real and imaginary part.

$$\int_{-\infty}^\infty\frac{\cos({k\sqrt{x^2+a^2})}}{\sqrt{x^2+a^2}}dx-i\int_{-\infty}^\infty\frac{\sin({k\sqrt{x^2+a^2})}}{\sqrt{x^2+a^2}}dx$$

And I tried to contour integral using branch cut similar to the link. However I can't handle branch cuts on the imaginary line well.(Is the answer of the link only way to handle it?) How to handle that branch cuts simple as the case of real line.

And.... How to evaluate the whole integral??

Thanks for several answers. I've thought the problem too hard..

However, I made a mistake to my question. I need to consider the case when $a=0$. I guess...

$$\int_{-\infty}^\infty\frac{e^{-ik\sqrt{x^2}}}{\sqrt{x^2}}dx=2\int_{0}^\infty\frac{e^{-ik\sqrt{x^2}}}{\sqrt{x^2}}dx=2\int_{0}^\infty\frac{e^{-ikx}}{x}dx$$ Is it valid?

Community
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SWnam
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2 Answers2

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Your integrand is even, so rewrite as $$ \int_{-\infty}^\infty\frac{e^{-ik\sqrt{x^2+a^2}}}{\sqrt{x^2+a^2}}dx=2\int_{0}^\infty\frac{e^{-ik\sqrt{x^2+a^2}}}{\sqrt{x^2+a^2}}dx\ , $$ then set $\sqrt{x^2+a^2}=t\Rightarrow x=\sqrt{t^2-a^2}$, yielding $$ 2\int_{a}^\infty\frac{e^{-ikt}}{t}\frac{2t dt}{2\sqrt{t^2-a^2}}=2\int_a^\infty\frac{e^{-i k t}dt}{\sqrt{t^2-a^2}}=2\int_1^\infty\frac{e^{-i k a t}dt}{\sqrt{t^2-1}}\ .$$ Mathematica can handle the resulting integral and gives the answer $2K_0(i a k)$, where $K_0$ is a Bessel function [checked numerically]. I'll try to see if I can find this integral on Gradshteyn.

EDIT The last integral is basically one of the integral representation of the Bessel function, see http://functions.wolfram.com/Bessel-TypeFunctions/BesselK/07/01/01/0001/

Pierpaolo Vivo
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THIS IS NOT AN ANSWER

putting the substitution $x = a\tan(\theta)$ greatly simplifies the integral. Thereafter, multiplying and dividing by $\tan(\theta)$ in the transformed integral, and again integrating by parts two times would give you the answer.

This may look a little lengthy but actually isn't.

Cameron Williams
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user317698
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  • If that is not an answer, you have to post is as a comment. Also, you should learn how to type in LaTeX. It will clarify ways better what you are meaning! Besides this, are you sure about the substitution? – Enrico M. Mar 02 '16 at 16:17
  • @1 over 137 err..you have messed up with the substitution...it is x=atan(theta) NOT arctan(theta) yes i am sure about the substitution.And i am learning how to type in LaTex..besides thanks for the editing – user317698 Mar 02 '16 at 16:20
  • Good then! ^^ If you can, you may try to write a complete answer! It would be useful and great! – Enrico M. Mar 02 '16 at 16:20
  • What is the difference between atan and arctan? – Pierpaolo Vivo Mar 02 '16 at 18:11
  • @PierpaoloVivo S/he meant $a\cdot\tan\theta$. – Cameron Williams Mar 03 '16 at 02:26