2
$\begingroup$

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

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

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


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

  • 0
    @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. – 2012-01-20

1 Answers 1

3

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.