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I have found the following Laplace Transform in a list

$$\int\limits_0^{\infty}e^{-st}\frac{e^{-u^2/4t}}{\sqrt{\pi t}}dt = \frac{e^{-u\sqrt{s}}}{\sqrt{s}}.$$

I am wondering how to prove this? I tried to do some substitutions for the integral, but nothing worked. Can someone explain it to me?

I appreciate any help!

Hasti Musti
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  • You could maybe use the fact that $\mathcal{L}^{-1}{F(s)G(s)}=(f*g)(t)$, setting $F(s) = e^{-u\sqrt{s}}=\mathcal{L}^{-1}{f(t)}$ and $G(s)=\sqrt{s}=\mathcal{L}^{-1}{g(t)}$. – giobrach Jun 22 '16 at 17:56
  • Hint: 10.32.10 here: http://dlmf.nist.gov/10.32 – tired Jun 22 '16 at 18:03

2 Answers2

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First, enforcing the substitution $t\to t^2$ in the integral of interest yields

$$\begin{align} \int_0^\infty e^{-st}\,\frac{e^{-u^2/4t}}{\sqrt{\pi t}}\,dt&=\frac{2}{\sqrt\pi}\int_0^\infty e^{-s\left(t^2+u^2/4st^2\right)}\,dt\\\\ &=\frac{2e^{-u/\sqrt{s}}}{\sqrt\pi}\int_0^\infty e^{-s\left(t-u/2\sqrt{s}t\right)^2}\,dt \tag 1 \end{align}$$

Second, enforcing the substitution $t\to \sqrt{\frac{u}{2\sqrt{s}}}\,t$ into the right-hand side of $(1)$ and letting $\alpha =\sqrt{\frac{u}{2\sqrt{s}}}$ reveals

$$\frac{2e^{-u/\sqrt{s}}}{\sqrt\pi}\int_0^\infty e^{-s\left(t-u/2\sqrt{s}t\right)^2}\,dt=\frac{2e^{-u\sqrt{s}}}{\sqrt\pi}\alpha \int_0^\infty e^{-\alpha^2 s\left(t-1/t\right)^2}\,dt \tag 2$$

Third, enforcing the substitution $t\to 1/t$ in the integral on the right-hand side of $(3)$, we obtain

$$\frac{2e^{-u\sqrt{s}}}{\sqrt\pi}\alpha \int_0^\infty e^{-\alpha^2 s\left(t-1/t\right)^2}\,dt =\frac{2e^{-u\sqrt{s}}}{\sqrt\pi}\alpha \int_0^\infty e^{-\alpha^2 s\left(t-1/t\right)^2}\,\left(\frac{1}{t^2}\right)\,dt \tag 3$$

Next, adding $(2)$ and $(3)$ and dividing by $2$, we find that

$$\begin{align} \frac{2e^{-u\sqrt{s}}}{\sqrt\pi}\alpha \int_0^\infty e^{-\alpha^2 s\left(t-1/t\right)^2}\,dt&=\frac{e^{-u\sqrt{s}}}{\sqrt\pi}\alpha \int_0^\infty e^{-\alpha^2 s\left(t-1/t\right)^2}\,\left(1+\frac{1}{t^2}\right)\,dt\\\\ &=\frac{e^{-u\sqrt{s}}}{\sqrt\pi}\alpha \int_{-\infty}^{\infty}e^{-\alpha^2 s u^2}\,du\\\\ &=\frac{e^{-u\sqrt{s}}}{\sqrt{s}} \tag 4 \end{align}$$

Finally, substituting $(4)$ into $(1)$ we find

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty e^{-st}\,\frac{e^{-u^2/4t}}{\sqrt{\pi t}}\,dt=\frac{e^{-u\sqrt{s}}}{\sqrt{s}}}$$

as was to be shown!

Mark Viola
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  • This is really awesome. How can you manage to find all those well fitting substitutions? I feel always lucky if I can guess any substitution which works fine, let alone it feels impossible to find so many correct substitutions in a row like you have done above...:-) – Hasti Musti Jun 23 '16 at 04:45
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    Thank you! The first substitution $t\to t^2$ gets rid of the square root term in the denominator. The second one normalizes the integration dummy variable and makes the exponent devoid of parameters. And the third $t\to 1/t$, exploits the fact that the exponent is invariant to this transformation. – Mark Viola Jun 23 '16 at 13:36
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It is probably easier to prove first that if $A,B\in\mathbb{R}^+$ we have: $$ \int_{0}^{+\infty}\exp\left(-A^2 x^2-\frac{B^2}{x^2}\right)\,dx = \frac{\sqrt{\pi}}{2A} e^{-2AB}\tag{1} $$ that is the same as proving that for every $C\in\mathbb{R}^+$ we have: $$ \int_{0}^{+\infty}\exp\left(-x^2-\frac{C^2}{x^2}\right)\,dx = \frac{\sqrt{\pi}}{2}e^{-2C}\tag{2} $$ or: $$ \int_{0}^{+\infty}\exp\left[-\left(x-\frac{C}{x}\right)^2\right]\,dx = \frac{\sqrt{\pi}}{2}.\tag{3} $$ However, $(3)$ is a trivial consequence of Glasser's master theorem.

So you just have to use the substitution $t=v^2$, then exploit $(1)$.

Jack D'Aurizio
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