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When I use the convolution method, I can't avoid getting a divergent integral.

Gregory Grant
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bananana
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  • http://www.wolframalpha.com/input/?i=inverse+laplace+%281%2F%28sqrt%7Bs%7D%28s-a%29%29 . Interesting thing is wolfram alpha shows this as a kind of shifted error function. – Someone May 17 '15 at 07:48
  • I was able to forward transformed the answer, but I'd be interested in the inverse transform as well. – Gappy Hilmore May 17 '15 at 22:30
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    http://math.stackexchange.com/questions/1009697/inverse-laplace-transform-of-e-c-sqrts-sqrtsa-s/1010327#1010327 – Ron Gordon May 18 '15 at 20:50

1 Answers1

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I will solve this using convolution but I would LOVE to see someone do it using contour integration.

Using the fact that:

$\frac{1}{\sqrt t}\rightarrow \sqrt \frac{\pi}{s}$ which means $\frac{1}{\sqrt s}\rightarrow \sqrt \frac{1}{\pi t}$

$\frac{1}{s-a}\rightarrow e^{at}u(t)$, where $u(t)$ is the unit step (which means the exponential is zero for $t<0$, which is probably where you went wrong)

We need to find:

$$\sqrt \frac{1}{\pi t} * e^{at}=\frac {1}{\sqrt \pi}\int ^t_0\frac{e^{(at-a\tau)}}{\sqrt \tau}d\tau$$

The reason the integral is from $0$ to $\infty$ is because the exponential $e^{-at}$ is zero for $t<0$. If you have trouble visualizing, here is a gif. The exponential we want to convolve looks like this.

$$=\frac {1}{\sqrt \pi}e^{at}\int ^t_0\frac{e^{-a\tau}}{\sqrt \tau}d\tau$$

substitute $a\tau=b^2$, we have:

$$=\frac {1}{\sqrt \pi}e^{at}\int ^{\sqrt{at}}_0 \frac{2}{\sqrt a}{e^{-b^2}}db$$

$$=\frac{1}{\sqrt a}e^{at}\frac {2}{\sqrt \pi}\int ^{\sqrt{at}}_0 {e^{-b^2}}db$$

$$=\frac{1}{\sqrt a}e^{at}erf(\sqrt{at})$$

Which matches wolfram's answer. Again, I'd love to see someone do this using contour integration.

Gappy Hilmore
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