How Do You Solve This Complex Bessel Function Integral?

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\int^{2a}_{0}dpJ[0,b\sqrt{p}]J[0,b\sqrt{2a-p}]

where a and b are constant, and J[0,x] is Bessel function.
 
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xylai said:
\int^{2a}_{0}dpJ[0,b\sqrt{p}]J[0,b\sqrt{2a-p}]

where a and b are constant, and J[0,x] is Bessel function.

Expand out both Bessel functions as a series and multiply them according to Cauchy's rule, namely

\left(\sum_{k=0}^{\infty}a_{k} x^{k}\right) \left(\sum_{j=0}^{\infty}b_{j} x^{j}\right) = \sum_{k=0}^{\infty}\sum_{j=0}^{k}a_{j}b_{k-j} x^k​

and pass the integral through to the inner most sum (the second time I've blatantly ignored convergence issues, perfering to hand-wave such until I get a result) and make the change of variables p=2au, you should get it from there... post your result so I can compare/check my work.

Ben
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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