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 September 15th, 2014, 03:05 AM #1 Newbie   Joined: Sep 2014 From: Adelaide Posts: 1 Thanks: 0 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! Thanks
 September 17th, 2014, 02:46 PM #2 Global Moderator   Joined: Dec 2006 Posts: 21,029 Thanks: 2259 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.
 September 29th, 2014, 09:22 AM #3 Senior Member     Joined: Dec 2013 From: some subspace Posts: 212 Thanks: 72 Math Focus: real analysis, vector analysis, numerical analysis, discrete mathematics $\displaystyle \mathcal{L} \left\{ \frac{1}{\sqrt{1-e^{-t}}} \right\} = \int_0 ^{\infty} \frac{e^{-st} \mathrm{d}t}{\sqrt{1-e^{-t}}}.$ 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: $\displaystyle \mathcal{L} \left\{ \frac{1}{\sqrt{1-e^{-t}}} \right\} = 2\int_0 ^{\frac{\pi}{2}} \cos ^{2s-1} \varphi \, \mathrm{d}\varphi = 2\int_0 ^{\frac{\pi}{2}} \cos ^{2s-1} \varphi \, \sin ^{2\cdot \frac{1}{2} -1} \varphi \, \mathrm{d}\varphi = \beta \left( s, \frac{1}{2} \right) = \frac{\sqrt{\pi} \, \Gamma \left( s \right)}{\Gamma \left( s + \frac{1}{2} \right)}$. Thanks from ZardoZ and jks
 September 29th, 2014, 03:42 PM #4 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus @fysmat nice! Thanks from fysmat

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