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 correspond 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?

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?

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 correspond 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?

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 correspond 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?