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I would appreciate if you could help me to find the following integral:

$$f(u)= \int_{-\infty }^{\infty} \frac{e^{-itu}}{{\sqrt {1+t^2}}} \;dt$$

Andrew D
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May
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1 Answers1

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This is the Modified Bessel Function: $2K_0(|u|)$.

Let $t=\sinh(x)$ $$ \begin{align} \int_{-\infty}^\infty\frac{e^{-itu}}{\sqrt{1+t^2}}\mathrm{d}t &=\int_{-\infty}^\infty e^{-iu\sinh(x)}\,\mathrm{d}x\\ &=\int_{-\infty}^\infty \cos(u\sinh(x))\,\mathrm{d}x\\[6pt] &=2K_0(|u|) \end{align} $$

robjohn
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  • Thank you, but I have a question; I find $K_0(u)$ in math world: http://mathworld.wolfram.com/ModifiedBesselFunctionoftheSecondKind.html so $K_0(|u|)=\int_{-\infty }^{\infty} cos(ut)/ \sqrt{1+t^2} $ what about $ \int_{-\infty }^{\infty} i sin(ut)/ \sqrt{1+t^2}$? – May Sep 16 '13 at 19:20
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    @May: $\frac{\sin(ut)}{\sqrt{1+t^2}}$ is an odd function of $t$, so the positive and negative reals will cancel. – robjohn Sep 16 '13 at 19:27
  • I have another question which somehow relates to this one. In this link http://mathworld.wolfram.com/NormalProductDistribution.html there is a PDF based on modified bessel function. but this PDF is for the case that both mean values are zero. Now I want to find the general formula for PDF of XY with non-zero means. I would appreciate if you could help me. Thank you. – May Sep 16 '13 at 19:37
  • @May: the best thing to do is to post a question, showing what you've gotten so far or what you've tried. It is hard to answer a question like that in a comment, and a larger group of people, who might know more about these things, would be able to answer. – robjohn Sep 16 '13 at 20:51
  • thank you for your suggestion I posted it http://math.stackexchange.com/questions/495773/how-to-calculate-the-following-double-integral – May Sep 16 '13 at 21:04