# Math Help - Proof!

1. ## Proof!

Prove that:

$\int_{0}^{\infty}(\sinh{x})^{\alpha-1}P_{v}^{-\mu}(\cosh{x})\,dx = \frac{2^{-1-\mu}\Gamma\left(\frac{1}{2}\alpha+\frac{1}{2}\mu\r ight)\Gamma\left(\frac{1}{2}v-\frac{1}{2}\alpha+1\right)\Gamma\left(\frac{1}{2}-\frac{1}{2}\alpha-\frac{1}{2}v\right)}{\Gamma\left(\frac{1}{2}\mu+\f rac{1}{2}v+1\right)\Gamma\left(\frac{1}{2}+\frac{1 }{2}\mu-\frac{1}{2}v\right)\Gamma\left(1+\frac{1}{2}\mu-\frac{1}{2}\alpha\right)}$

Code:


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$\text{Re} \left(\alpha+\mu\right) > 0$,

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$\text{Re} \left(v-\alpha+2\right) > 0$,

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$\text{Re} (1-\alpha-v) > 0$.

2. Originally Posted by Legendre
Prove that:

$\int_{0}^{\infty}(\sinh{x})^{\alpha-1}P_{v}^{-\mu}(\cosh{x})\,dx = \frac{2^{-1-\mu}\Gamma\left(\frac{1}{2}\alpha+\frac{1}{2}\mu\r ight)\Gamma\left(\frac{1}{2}v-\frac{1}{2}\alpha+1\right)\Gamma\left(\frac{1}{2}-\frac{1}{2}\alpha-\frac{1}{2}v\right)}{\Gamma\left(\frac{1}{2}\mu+\f rac{1}{2}v+1\right)\Gamma\left(\frac{1}{2}+\frac{1 }{2}\mu-\frac{1}{2}v\right)\Gamma\left(1+\frac{1}{2}\mu-\frac{1}{2}\alpha\right)}$

Code:


img.top {vertical-align:15%;}

$\text{Re} \left(\alpha+\mu\right) > 0$,

img.top {vertical-align:15%;}

$\text{Re} \left(v-\alpha+2\right) > 0$,

img.top {vertical-align:15%;}

$\text{Re} (1-\alpha-v) > 0$.
Challenge problems should be something that requires you to be clever, nto something that takes two weeks to do because it is so dang laborious. Does this have a clever solution or did you just find the most complicated looking integral on the internet and post it up here?

3. Originally Posted by Drexel28
Challenge problems should be something that requires you to be clever, nto something that takes two weeks to do because it is so dang laborious. Does this have a clever solution or did you just find the most complicated looking integral on the internet and post it up here?
dang laborious!

4. Originally Posted by Drexel28
Challenge problems should be something that requires you to be clever, nto something that takes two weeks to do because it is so dang laborious. Does this have a clever solution or did you just find the most complicated looking integral on the internet and post it up here?
xD