Prove in detail that (1/√t)[convolution](1/√t)=π URGENT DIFFERENTIAL EQUATION QUESTION!

Prove in detail that (1/√t)[convolution](1/√t)=π
10 points if you can solve by 3:30 central time, 4/21/11

Possible help
F[convolution]G =
t
⌠ f(t-x)g(x)dx-->int[1/(√x) * 1/√(t-x) * dx] {from 0-->t}
⌡
0
-->[ 2arctan((√x)/√(t-x)) ] {from 0 to t} = sin(inf)*π =π

I don't know how to do the integral int[1/(√x) * 1/√(t-x))dx {from 0-->t} I just know what the answer should be....

∫(x = 0 to t) [(1/√x) 1/√(t - x)] dx
= ∫(x = 0 to t) dx / √[x(t - x)]
= ∫(x = 0 to t) dx / √(tx - x^2)
= ∫(x = 0 to t) dx / √(t^2/4 - t^2/4 + tx - x^2), completing the square
= ∫(x = 0 to t) dx / √[t^2/4 - (x - t/2)^2]
= arcsin [(x - t/2) / (t/2)] {for x = 0 to t}
= arcsin [(2x - t) / t] {for x = 0 to t}
= arcsin 1 - arcsin(-1)
= π.

I hope this helps!

Yes I am.

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