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Prove that

$$\int_{0}^{\infty} \sin \left(x\right) \sin \left(\frac{a}{x}\right) \ dx = \frac{\pi \sqrt{a}}{2} J_{1} \left( 2 \sqrt{a} \right)$$ where $J_{1}$ is the Bessel function of the first kind of order 1.

Some calculations I have done

$$\int_{0}^{\infty} \sin \left(x\right) \sin \left(\frac{a}{x}\right) \ dx= \int_{0}^{\infty} \sum_{k=0}^{\infty }(-1)^{k}\frac{x^{2k+1}}{2k+1!} \cdot \sum_{l=0}^{\infty }(-1)^{l}\frac{a^{2l+1}x^{-2l-1}}{2l+1!} \ dx$$

$$= \int_{0}^{\infty} \sum_{l=0}^{\infty } \sum_{k=0}^{\infty }(-1)^{k+l}\frac{x^{2(k-l)}}{(2k+1)!(2l+1)!} a^{2l+1} \ dx$$


$$\frac{\pi \sqrt{a}}{2}J_{1}(2sqrt{a})=\frac{\pi \sqrt{a}}{2} \sum_{l=0}^{\infty}\frac{(-1)^l}{2^{2l+1}l!(1+l)!} 2^{l+\frac{1}{2}}a^{l+\frac{1}{2}}$$ $$=\pi \sum_{l=0}^{\infty}\frac{(-1)^l}{2^{l+\frac{3}{2}}l!(1+l)!} a^{l+1}$$

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You haven't asked any question. –  joriki Jan 18 '12 at 17:29
    
added question + some calculations –  wnvl Jan 20 '12 at 0:39
    
@AlexBecker, your comment is considerably cryptic. Could you add another one which is more obviously helpful? –  Mariano Suárez-Alvarez Jan 20 '12 at 20:03
    
@MarianoSuárez-Alvarez The original question was just the equality $$\int_{0}^{\infty} \sin \left(x\right) \sin \left(\frac{a}{x}\right) \ dx = \frac{\pi \sqrt{a}}{2} J_{1} \left( 2 \sqrt{a} \right)$$ with no words or indication of what he wanted, so I left that comment as an illustration of how hard to understand such a question is. Since the question has been edited, I have deleted my comment. –  Alex Becker Jan 20 '12 at 20:56

1 Answer 1

up vote 3 down vote accepted

Let $f(a) = \int_0^\infty \sin(x) \sin\left(\frac{a}{x}\right) \mathrm{d} x$, where $a\in \mathbb{R}$. Without loss of generality we can assume $a > 0$.

Observe that $$ \begin{eqnarray} a f^{\prime\prime}(a) = \int_0^\infty \sin(x) \left(-\sin\left(\frac{a}{x} \right) \right) \frac{a}{x} \frac{\mathrm{d} x}{x} \stackrel{x \to a/y}{=} \int_0^\infty \sin\left( \frac{a}{y} \right) (-\sin(y)) \mathrm{d} y = -f(a) \end{eqnarray} $$ The differential equation so obtained, $a f^{\prime\prime}(a) + f(a) = 0$, reduces to Bessel differential equation, with general solution $$ f(a) = c_1 \cdot \sqrt{a} J_1(2\sqrt{a}) + c_2 \cdot \sqrt{a} Y_1(2\sqrt{a}) $$ where $J_1$ and $Y_1$ are Bessel functions of the first and the second kind. Since $f(0)=0$ and $\lim_{a \downarrow 0^+} \sqrt{a} Y_1(2 \sqrt{a} ) = -\frac{1}{\pi}$, we get $c_2 = 0$.

By splitting the integration range into $(0,1)$ and $(1,\infty)$ and changing variables in the first one to $x \to 1/x$ we get $$ f(a) = \int_{1}^{\infty} \sin(1/x) \sin(a x) \frac{\mathrm{d} x}{x^2} + \int_1^\infty \sin(x) \sin(a/x) \mathrm{d} x $$ From here we see that small $a$ expansion first term is $$ f(a) = a \left( \int_{1}^\infty \frac{1}{x} \sin(x) \mathrm{d} x + \int_{1}^\infty \sin(1/x) \frac{\mathrm{d} x }{x} \right) + \mathcal{o}(a) = \frac{\pi}{2} a + \mathcal{o}(a) $$ This fixes $c_1 = \frac{\pi}{2}$, proving the requested equality.

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