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I had this on a test today and can't get it out of my head.
I want to know if I got it right or wrong.
The question was to find the inverse Laplace transformation for the following:
$$\frac{\sqrt{\pi}}{\sqrt{s-3}}$$
My answer was $$\frac{1}{\sqrt{e^{3t}}}$$
However the more I think about it, I'm beginning to think it should be $$\sqrt{e^{3t}}$$
Please let me know which is right (if either of them).
Ok, so now I know I was incorrect in both cases. Now I want to understand where I went wrong in my process. I was using the the fact that the inverse transformation of $$\frac{\sqrt{\pi}}{\sqrt{s}}$$ is $$\frac{1}{\sqrt{t}}$$ Which would leave me with $$\frac{1}{\sqrt{t-3}}$$
Then I decided I'd apply the fact that the inverse transformation of $$\frac{1}{s-a}$$ is $$e^{at}$$
Which lead me to think $$\frac{1}{\sqrt{t-3}} = \sqrt{\frac{1}{t-3}} = \sqrt{e^{3t}} $$
NOTE
I'm not looking for another way to figure this inverse transformation out (after all someone has already posted the answer), I'm simply wanting to understand where I went wrong so I do not make this mistake again in the future (like on my final).

I figured it out! See answer below.

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Both are wrong... –  Fabian Apr 15 '11 at 20:34
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A crazy idea, but: have you thought about talking to your instructor? –  Pete L. Clark Apr 15 '11 at 22:57
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@Pete L. Clark Well if you read my post, you would know I just had the test and will not see my instructor until Monday. I also stated that the problem was bothering me because I apparently do not understand it. Hence, why I thought I'd post it and see if anyone could shed some light. –  enlgmatlc Apr 16 '11 at 1:42

2 Answers 2

You know that the inverse transform of $\sqrt{\pi}/\sqrt{s}$ is $1/\sqrt{t}$. Futhermore, you (hopefully) know that if you shift $s\mapsto s-a$ the original function gets multiplied by $e^{a t}$. Putting this two results together, you obtain that the inverse transform of $\sqrt{\pi}/\sqrt{s-3}$ is given by $$ \frac{e^{3t}}{\sqrt{t}}.$$

(the remainder is an answer to the question before it got edited and most probably does not help the OP)

The inverse of the Laplace transform is given by the Bromwich integral $$f(t) = \frac{1}{2\pi i} \int_C ds \, e^{st} \frac{\sqrt{\pi}}{\sqrt{s-3}}$$ where $C$ is a contour in the complex plane with the real part larger than any singularity (here we have a branch point at $s=3$. We deform the contour such that we start at $s=-\infty -i 0^+$ the go along the branch cut of $\sqrt{s-3}$ (which we chose along the real line from $s=3$ to $s=-\infty$) up to $s=3$ and back on the other side of the branch cut to $s=-\infty +i0^+$. The back transform therefore assume the form $$ \begin{align} f(t)&= \frac{1}{2\sqrt{\pi} i} \int_{-\infty}^3 ds\, e^{st} \left[ \frac{1}{\sqrt{s-3-i0^+}} - \frac{1}{\sqrt{s-3+i0^+}} \right] \\ &=\frac{e^{3 t}}{\sqrt{\pi} } \int_0^\infty d\tilde{s}\, \frac{e^{-\tilde{s}t} } {\sqrt{\tilde{s}}} = \frac{e^{3 t}}{\sqrt{\pi} \sqrt{t} } \int_0^\infty dx\, \frac{e^{- x }} {\sqrt{x}}\\ &=\frac{e^{3 t}}{ \sqrt{t} } \end{align},$$ where $s= 3-\tilde{s}$ and $t>0$ (of course the final result can be analytically continued).

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To solve these problems we have not been using the Bromwich integral, my professor has actually been having us use tables of basic facts about Laplace Transformations to figure out the inverse transformations. Seeing you use this integral to prove the answer makes me glad my professor isn't having us derive the answers by definition as you did. It looks kinda painful. :P –  enlgmatlc Apr 16 '11 at 3:43
up vote 2 down vote accepted

I figured it out!
I see why both my answers are wrong. If I would have just thought to check my answer by taking the Laplace Transformation of my answers to verify whether or not they were correct, I (hopefully) would have observed that I needed $e^{3t}$ in the numerator instead of the denominator. Which would have lead me to think I should have $\sqrt{t}$ in the denominator. Special thanks to Fabian.

What I did wrong
In my first step, I was correct in applying what I knew about the fact that the Laplace transform of $$\frac{1}{\sqrt{t}} = \frac{\sqrt{\pi}}{\sqrt{s}}.$$ Everything after this point is where I went wrong. I should have then applied the fact that the Laplace transform of $$e^{at}f(t) = F(s-a).$$ Which would have lead me to $$\frac{e^{3t}}{\sqrt{t}}.$$
This is so disappointing! I lost ten points on my test due not fully analyzing the problem!!

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