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September 15th, 2014, 03:05 AM   #1
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Laplace Transform

I am usually not bad with laplace transform questions, but have not had any luck with this question:
Find the laplace transform of (1-exp(-t))^(-1/2)
Just any hints on where to start would be great!
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September 17th, 2014, 02:46 PM   #2
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According to W|A, it's $\displaystyle \frac{\sqrt\pi\Gamma(s)}{\Gamma(s+\frac12)}$. I'm not sure how to prove that by elementary means.
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September 29th, 2014, 09:22 AM   #3
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$\displaystyle \mathcal{L} \left\{ \frac{1}{\sqrt{1-e^{-t}}} \right\} = \int_0 ^{\infty} \frac{e^{-st} \mathrm{d}t}{\sqrt{1-e^{-t}}}.$

Let's substitute $\displaystyle 1-e^{-t} = x^2 \quad \Rightarrow \quad \mathrm{d}t = \frac{2x\mathrm{d}x}{1-x^2}$.

One gets

$\displaystyle \mathcal{L} \left\{ \frac{1}{\sqrt{1-e^{-t}}} \right\} = 2\int_0 ^{1} \left( 1 - x^2 \right) ^{s - 1} \mathrm{d}x$.

Then, after substituting $\displaystyle x = \sin \varphi$, the integral can be expressed in terms of beta function:

$\displaystyle \mathcal{L} \left\{ \frac{1}{\sqrt{1-e^{-t}}} \right\} = 2\int_0 ^{\frac{\pi}{2}} \cos ^{2s-1} \varphi \, \mathrm{d}\varphi = 2\int_0 ^{\frac{\pi}{2}} \cos ^{2s-1} \varphi \, \sin ^{2\cdot \frac{1}{2} -1} \varphi \, \mathrm{d}\varphi = \beta \left( s, \frac{1}{2} \right) = \frac{\sqrt{\pi} \, \Gamma \left( s \right)}{\Gamma \left( s + \frac{1}{2} \right)}$.
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September 29th, 2014, 03:42 PM   #4
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@fysmat nice!
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