Laplace transform of function

Question

Assume that $f(u)=(\frac{b}{πu^3})^{1/2} e^{2b} e^{-bu} e^{-b/u}$, where $b>0.$ I am trying to calculate the Laplace transform $L\{f(u)\}(s)$ and then the $n_{th}$ derivative of this transform, $L^{n}(s),$ but i have difficulties in calculating the integrals, especially in the second one. Can someone help in both of these integrals?

Answer 1

So you want to evaluate the integral

$$I = \int_0^{\infty} du \, u^{-3/2} e^{-(a u + b/u)} $$

where $a=s+b$. Let $u = v^2$; then

$$I = 2 \int_0^{\infty} \frac{dv}{v^2} e^{-(a v^2+b/v^2)} = 2 e^{2 \sqrt{a b}} \int_0^{\infty} \frac{dv}{v^2} e^{-(\sqrt{a} v+\sqrt{b}/v)^2}$$

Now let $y=\sqrt{a} v+\sqrt{b}/v$. Then

$$v = \frac{y\pm \sqrt{y^2-4 \sqrt{a b}}}{2 \sqrt{a}} $$ $$dv = \frac1{2 \sqrt{a}} \left (1 \pm \frac{y}{\sqrt{y^2-4 \sqrt{a b}}} \right ) dy $$ $$\frac1{v} = \frac{y\mp \sqrt{y^2-4\sqrt{a b}}}{2 \sqrt{b}} $$

Then the integral is

$$I = \frac{e^{2 \sqrt{a b}}}{4 \sqrt{a b^2}} \int_{\infty}^{2 (a b)^{1/4}} dy \left (1 - \frac{y}{\sqrt{y^2-4 \sqrt{a b}}} \right ) \left ( y+ \sqrt{y^2-4 \sqrt{a b}} \right )^2 e^{-y^2} \\ + \frac{e^{2 \sqrt{a b}}}{4 \sqrt{a b^2}} \int_{2 (a b)^{1/4}}^{\infty} dy \left (1 + \frac{y}{\sqrt{y^2-4 \sqrt{a b}}} \right ) \left ( y- \sqrt{y^2-4 \sqrt{a b}} \right )^2 e^{-y^2} $$

The splitting up of the integral into the above two pieces corresponds to the two branches of $v(y)$ defined above. We can simplify somewhat to obtain

$$\begin{align}I &= \frac{e^{2 \sqrt{a b}}}{2 \sqrt{a b^2}} \left [\int_{2 (a b)^{1/4}}^{\infty} dy \frac{y^3}{\sqrt{y^2-4 \sqrt{a b}}} e^{-y^2} - \int_{2 (a b)^{1/4}}^{\infty} dy\, y \sqrt{y^2-4 \sqrt{a b}} e^{-y^2} \right ] \\ &= \frac{e^{2 \sqrt{a b}}}{4 \sqrt{a b^2}} \left [\int_{4 \sqrt{a b}}^{\infty} dy \frac{y}{\sqrt{y-4 \sqrt{a b}}} e^{-y} - \int_{4 \sqrt{a b}}^{\infty} dy\, \sqrt{y-4 \sqrt{a b}} e^{-y} \right ]\\ &= \frac{e^{-2 \sqrt{a b}}}{\sqrt{b}} \int_0^{\infty} dy \, y^{-1/2} e^{-y}\\ &= \sqrt{\frac{\pi}{b}} e^{-2 \sqrt{a b}}\end{align}$$

To directly address your transform, just plug in $a=b+s$ and using your $f$, we get

$$F(s) = \int_0^{\infty} du \, f(u) e^{-s u} = e^{-2 \sqrt{b} (\sqrt{b+s}-\sqrt{b})} $$