# Proof!

• Dec 16th 2009, 07:34 PM
Legendre
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:

 <br /> img.top {vertical-align:15%;}<br /> $\text{Re} \left(\alpha+\mu\right) > 0$, <br /> img.top {vertical-align:15%;}<br /> $\text{Re} \left(v-\alpha+2\right) > 0$, <br /> img.top {vertical-align:15%;}<br /> $\text{Re} (1-\alpha-v) > 0$.
• Dec 16th 2009, 09:46 PM
Drexel28
Quote:

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:

 <br /> img.top {vertical-align:15%;}<br /> $\text{Re} \left(\alpha+\mu\right) > 0$, <br /> img.top {vertical-align:15%;}<br /> $\text{Re} \left(v-\alpha+2\right) > 0$, <br /> img.top {vertical-align:15%;}<br /> $\text{Re} (1-\alpha-v) > 0$.

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?
• Dec 17th 2009, 12:53 AM
NonCommAlg
Quote:

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! (Rofl)
• Jan 7th 2010, 03:20 PM
TWiX
Quote:

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
(Rofl)(Rofl)