Mostre que
\int_{0}^{\infty} J_{\mu}(x) J_{\nu}(x) \frac{d x}{x}=\frac{2}{\pi} \frac{\operatorname{sen}[(\mu-\nu) \pi / 2]}{\mu^{2}-\nu^{2}}, \quad \mu+\nu>-1 .

Passo 1

O enunciado pede para mostrar que, sob a condição \mu+\nu>-1, vale
\int_{0}^{\infty}J_{\mu}(x)\,J_{\nu}(x)\,\frac{dx}{x} =\frac{2}{\pi}\,\frac{\sin\!\bigl[(\mu-\nu)\pi/2\bigr]}{\mu^{2}-\nu^{2}}.
O conteúdo necessário é o chamado integral de Weber–Schafheitlin para produtos de funções de Bessel. Vamos usar a fórmula geral e, em seguida, especializá-la a \lambda=1.

Passo 2

A fórmula de Weber–Schafheitlin (ver, por exemplo, Gradshteyn–Ryzhik 6.574.2) afirma que, para
\Re(\mu+\nu+1)>\Re(\lambda)>0,
\int_{0}^{\infty}x^{-\lambda}J_{\mu}(x)\,J_{\nu}(x)\,dx =\frac{\Gamma(\lambda)\,\Gamma\!\Bigl(\tfrac{\mu+\nu+1-\lambda}{2}\Bigr)} {2^{\lambda}\, \Gamma\!\Bigl(\tfrac{\nu-\mu+1+\lambda}{2}\Bigr)\, \Gamma\!\Bigl(\tfrac{\mu-\nu+1+\lambda}{2}\Bigr)\, \Gamma\!\Bigl(\tfrac{\mu+\nu+1+\lambda}{2}\Bigr)}.

Passo 3

Neste problema temos \lambda=1. Logo
\int_{0}^{\infty}x^{-1}J_{\mu}(x)\,J_{\nu}(x)\,dx =\frac{\Gamma(1)\,\Gamma\!\bigl(\tfrac{\mu+\nu}{2}\bigr)} {2^{1}\, \Gamma\!\bigl(\tfrac{\nu-\mu+2}{2}\bigr)\, \Gamma\!\bigl(\tfrac{\mu-\nu+2}{2}\bigr)\, \Gamma\!\bigl(\tfrac{\mu+\nu+2}{2}\bigr)}.
Como \Gamma(1)=1, defina
A=\frac{\mu+\nu}{2}, \quad B=\frac{\mu-\nu}{2}.
Então
\int_{0}^{\infty}\frac{J_{\mu}(x)\,J_{\nu}(x)}{x}\,dx =\frac{\Gamma(A)}{2\;\Gamma(1-B)\,\Gamma(1+B)\,\Gamma(A+1)}.

Passo 4

Use a relação de recorrência \Gamma(A+1)=A\,\Gamma(A). Assim
\int_{0}^{\infty}\frac{J_{\mu}(x)\,J_{\nu}(x)}{x}\,dx =\frac{\Gamma(A)}{2\,\Gamma(1-B)\,\Gamma(1+B)\,A\,\Gamma(A)} =\frac{1}{2\,A\,\Gamma(1-B)\,\Gamma(1+B)}.

Passo 5

Para o produto \Gamma(1+B)\,\Gamma(1-B) usamos a fórmula de reflexão
\Gamma(1+B)\,\Gamma(1-B) = B\,\Gamma(B)\,\Gamma(1-B) = B\,\frac{\pi}{\sin(\pi B)}.
Logo
\int_{0}^{\infty}\frac{J_{\mu}(x)\,J_{\nu}(x)}{x}\,dx =\frac{1}{2\,A}\,\frac{\sin(\pi B)}{\pi\,B} =\frac{\sin\!\bigl(\pi\frac{\mu-\nu}{2}\bigr)}{2\,\pi\,\frac{\mu+\nu}{2}\,\frac{\mu-\nu}{2}} =\frac{2}{\pi}\,\frac{\sin\!\bigl[(\mu-\nu)\pi/2\bigr]}{\mu^{2}-\nu^{2}}.

Resposta

\boxed{ \int_{0}^{\infty}J_{\mu}(x)\,J_{\nu}(x)\,\frac{dx}{x} =\frac{2}{\pi}\,\frac{\sin\!\bigl[(\mu-\nu)\pi/2\bigr]}{\mu^{2}-\nu^{2}}, \quad \mu+\nu>-1. }

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