We now try to attack the following inverse LT:

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

I could not find any tables that had this paired with its ILT. It is also strange in that the branch point of the function is not at zero, as it was in the other 2 cases. Nevertheless, as will be seen below, the approach I take here doesn’t change from the approaches below.

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

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

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

Then

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

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][2]. Plug this expression in the equation for the ILT and you are done.

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