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I am a PhD student working on complex analysis. After integrating over a keyhole contour to obtain the inverse of a particular Laplace transform, I ended up with the following integral:

$$\frac{1}{\pi}\int^{4b}_0 e^{-tx}\biggl(\frac{\sqrt{4bx-x^2}}{(2b-2c)^2+4cx}\biggr) dx $$

where $b$, $c$ and $ t $ are positive constants. This integral corresponds to the linear segments of the contour, which has two branch points.

My current progress:

$$ \frac{1}{\pi}\int^{4b}_0 e^{-tx}\biggl(\frac{\sqrt{(2b)^2-(x-2b)^2}}{(2b-2c)^2+4cx}\biggr) dx. $$

Using trigonometric substitution: $ (x-2b) = 2b\sin z $

$$ \frac{1}{\pi}\int e^{-t(\ 2b+2b\sin z)}\biggl(\frac{4b^2\cos^2 z}{(2b-2c)^2 + 8bc\ +4c(2b\sin z)}\biggr) dz. $$

Could someone please help me to continue or show me a different way to approach to the problem?

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  • $\begingroup$ It might help to know what was the function whose inverse Laplace transform you started with. $\endgroup$
    – Noam D. Elkies
    Commented Feb 17, 2022 at 20:07
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    $\begingroup$ for $c=b$ your integral evaluates to $\frac{1}{2} \pi e^{-2 b t} [I_0(2 b t)+I_1(2 b t)]$ --- for $c\neq b$ a closed-form solution is not likely to be forthcoming. $\endgroup$
    – Carlo Beenakker
    Commented Feb 17, 2022 at 20:18

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