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I have an expression below which I need to do the laplace transform

Any help is highly appreicated.

The expression is :

$$ \frac{\exp\left(\frac{x}{2}\sqrt{(U/D)^2+4s/D}\right)}{s\sqrt{(U/D)^2+4s/D}} $$

The integral of Ron Gordon's expression:

The first expression on the solution was extracted from Wolfram from the following link: enter image description here

enter image description here

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The numerator is part of exponential – Jdbaba Apr 6 '13 at 1:47
I have edited to make that clearer to the naked eye. Also, your first sentence says you need to do the laplace transform, whereas the title says laplace inverse transform (and you have the inverse-problems tag); clarify if it's indeed the latter? – anon Apr 6 '13 at 2:18
up vote 2 down vote accepted

I will reduce this to an integral over the real line. Consider the integral in the complex plane:

$$\displaystyle \oint_C \frac{dz}{z} \frac{\exp{\left[\frac{x}{2} \sqrt{\left(\frac{U}{D} \right)^2 + 4 \frac{z}{D}}\right]}}{\sqrt{\left(\frac{U}{D} \right)^2 + 4 \frac{z}{D}}} e^{t z}$$

where $C$ is the following contour


Note the point is the branch point. I will assert without proof that the integral vanishes along the sections $C_2$, $C_4$, and $C_6$ of $C$. This leaves $C_1$ (the ILT), $C_3$, and $C_5$. There is a pole within the contour at $z=0$, so by the residue theorem, we have

$$\frac{1}{i 2 \pi}\left [\int_{C_1} + \int_{C_3} + \int_{C_5} \right ]\frac{dz}{z} \frac{\exp{\left[\frac{x}{2} \sqrt{\left(\frac{U}{D} \right)^2 + 4 \frac{z}{D}}\right]}}{\sqrt{\left(\frac{U}{D} \right)^2 + 4 \frac{z}{D}}} e^{t z} = \frac{D}{U} e^{U x/(2 D)}$$

Along $C_3$, note that there is a branch point at $z=-U^2/(4 D)$. Thus we parametrize $z=-U^2/(4 D) + e^{i \pi} y$; the integral over $C_3$ becomes

$$-i \int_{\infty}^0 \frac{dy}{y+\frac{U^2}{4 D}} \frac{e^{i (x/2) \sqrt{y}}}{\sqrt{y}} e^{-t y}$$

Similarly, along $C_5$, let $z=-U^2/(4 D) + e^{-i \pi} y$; the integral over $C_5$ becomes

$$i \int_0^{\infty} \frac{dy}{y+\frac{U^2}{4 D}} \frac{e^{-i (x/2) \sqrt{y}}}{\sqrt{y}} e^{-t y}$$

Putting this all together as above, we get an expression for the ILT:

$$\frac{1}{i 2 \pi}\int_{c-i \infty}^{c+i \infty} \frac{ds}{s} \frac{\exp{\left[\frac{x}{2} \sqrt{\left(\frac{U}{D} \right)^2 + 4 \frac{s}{D}}\right]}}{\sqrt{\left(\frac{U}{D} \right)^2 + 4 \frac{s}{D}}} e^{t s} = \\ \frac{D}{U} e^{U x/(2 D)} - \frac{1}{\pi} \int_0^{\infty} \frac{dy}{y+\frac{U^2}{4 D}} \frac{\cos{[(x/2) \sqrt{y}]}}{\sqrt{y}} e^{-t y}$$

So evaluation of the posted ILT depends on the ability to evaluate

$$\int_0^{\infty} \frac{dy}{y+\frac{U^2}{4 D}} \frac{\cos{[(x/2) \sqrt{y}]}}{\sqrt{y}} e^{-t y}$$

This integral may be evaluated first by substituting $y=u^2$ and then applying the convolution theorem (or Parseval's theorem, depending on your mood). The substitution produces

$$\int_{-\infty}^{\infty} \frac{du}{u^2+b^2} e^{-a u^2} e^{i k u}$$

where $a = t$, $b^2=U^2/(4 D)$, and $k=x/2$. You may then use the convolution theorem on the Fourier transforms of the functions

$$\int_{-\infty}^{\infty} e^{-a u^2} e^{i k u} = \sqrt{\frac{\pi}{a}} e^{-k^2/(4 a)}$$

$$\int_{-\infty}^{\infty} \frac{du}{u^2+b^2} e^{i k u} = \frac{\pi}{b} e^{-b |k|}$$


$$\int_{-\infty}^{\infty} \frac{du}{u^2+b^2} e^{-a u^2} e^{i k u} = \frac{1}{2 \pi} \sqrt{\frac{\pi}{a}} \frac{\pi}{b} \int_{-\infty}^{\infty} dk' e^{-(k-k')^2/(4 a)} e^{-b |k'|}$$

Frankly, the evaluation of this integral is straightforward but a mess, the derivation of which is not very instructive and will only serve to obfuscate the result. I leave it to the reader with assurances that I have done this out myself, by hand. The result is that

$$\int_0^{\infty} \frac{dy}{y+\frac{U^2}{4 D}} \frac{\cos{[(x/2) \sqrt{y}]}}{\sqrt{y}} e^{-t y} = \pi \frac{\sqrt{4 D}}{U} e^{U^2 t/(4 D)} \left [ e^{-U x/(4\sqrt{D})} \text{erfc}\left(-U \sqrt{\frac{t}{D}}+\frac{x}{4 \sqrt{t}}\right) + e^{U x/(4 \sqrt{D})} \text{erfc}\left(-U \sqrt{\frac{t}{D}}-\frac{x}{4 \sqrt{t}}\right) \right]$$

where erfc is the complementary error function. Plug this expression in the equation for the ILT and you are done.

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You are a genius of solving contour integrals! Can you also help me on e.g. expressing $\mathcal{L}^{-1}_{s\to x}\{e^{as^2+bs}\}$ $(a\neq0)$ in terms of integrals with real lower limit and upper limit in – doraemonpaul Apr 6 '13 at 16:02
@doraemonpaul: I am flattered, but I don't think I'd call myself that. In any case, I am not sure how these techniques could help you there, because that function has no poles or branch point singularities. Further, I am not sure of the convergence of such an integral over the contours I have laid out. Perhaps a wedge would work - let me think about it a bit. – Ron Gordon Apr 6 '13 at 16:14
@RonGordon Dear Ron, Thank you so much for your detailed solution. This is exactly what I needed. It took a while to get back to you because I was evaluating the values from the solution you gave. One thing I am not sure of is whether we have term "t" on the exponential of the part outside the large brackets in the final expression. I would appreciate if you could send me your hand made calculation to my email so that I could refer that as well. My email id is . Thanks again for your time and effort. You are a life saver. – Jdbaba Apr 7 '13 at 18:17
@Jdbaba: I will send you the hand calc, but it may be a little bit as work has caught up to me. I think the $t$ is right, but check my calc when you get it. I also compared with Mathematica output but the notation was not spot on. Also, please accept the solution if you find it useful. Thanks. – Ron Gordon Apr 7 '13 at 18:45
@RonGordon I have one more question to ask here. How would the solution change if I have negative sign inside the exponential in my original question ? – Jdbaba Apr 7 '13 at 18:49

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