trig integral with divergent pieces, but converges together

[galactus]

Show that:

$$\int_{0}^{\pi/2}\left[\left(\frac{3}{x^{2}}-1\right)\sin(x)-\frac{3}{x}\cos(x)\right]^{2}dx=\frac{\pi}{4}-\frac{24}{\pi^{3}}$$

[Hamza Mahmood]

We have that:

$$\int \left [ \left ( \frac{3}{x^2}-1 \right )\sin x - \frac{3}{x}\cos x \right ]^2 \, dx = \int \left ( \frac{9}{x^4}\sin^2 x +\sin^2 x - \frac{6}{x^2}\sin^2 x + \frac{9}{x^2}\cos^2 x -\frac{18}{x^3}\sin x \cos x + \frac{3}{x}\sin 2x \right )\, dx$$

Using Tabular Integration, we have that:

• $\displaystyle \int \frac{\sin^2 x}{x^4}\, dx = -\frac{1}{3x^3}\sin^2 x - \frac{1}{6x^2}\sin 2x - \frac{1}{3x}\cos 2x - \frac{2}{3}\int \frac{\sin 2x}{x}\, dx$
• $\displaystyle \int \frac{\sin^2 x}{x^2}\, dx=- \frac{\sin^2 x}{x} + \int \frac{\sin 2x}{x}\, dx$
• $\displaystyle \int \frac{\cos^2 x}{x^2}\, dx = - \frac{\cos^2 x}{x}- \int \frac{\sin 2x}{x}\, dx$
• $\displaystyle \int \frac{\sin x \cos x}{x^3}\, dx = - \frac{1}{4x^2}\sin 2x - \frac{1}{2x}\cos 2x - \int \frac{\sin 2x}{x}\, dx$

Putting them all together and simplifying we have that:

$$\int \left [ \left ( \frac{3}{x^2}-1 \right )\sin x - \frac{3}{x}\cos x \right ]^2 \, dx =\frac{x}{2}-\frac{3}{x}- \frac{\sin 2x}{4}- \frac{3}{x^3}\sin^2 x +\frac{3}{x^2}\sin 2x + \frac{3}{x}\sin^2 x+c$$

Plugging limits (and of course taking limits as $x \rightarrow 0+$) we get that:

$$\int_0^{\pi/2} \left [ \left ( \frac{3}{x^2}-1 \right )\sin x - \frac{3}{x}\cos x \right ]^2 \, dx =\frac{\pi}{4}- \frac{24}{\pi^3}$$


as claimed.

[r9m]

We might connect the given integrand to the spherical bessel function v.i.a. Rayleigh Formula:

$$j_n(x) = (-1)^nx^n \left(\frac{1}{x}\frac{\mathrm d}{\mathrm dx}\right)^n \frac{\sin x}{x}$$

and note that our case is $n = 2$, $\displaystyle j_2(x) = \left(\frac{3}{x^2} - 1\right)\frac{\sin x}{x} - \frac{3}{x^2}\cos x$

Thus, $$I = \int_0^{\pi/2} \left[\left(\frac{3}{x^2} - 1\right)\sin x - \frac{3}{x}\cos x\right]^2\,dx = \int_0^{\pi/2} x^2(j_2(x))^2\,dx$$

where, $\displaystyle j_n(x) = \sqrt{\frac{\pi}{2x}}J_{n+\frac{1}{2}}(x)$, ($J_{\nu}(x)$ being the $\nu$-th bessel function)

If we recall the differential equation of $\nu$-th bessel functions, it satisfies:

$$x^2y'' + xy' + (x^2-\nu^2)y = 0$$

If we multiply both sides with $2y'$ we can rewrite it as:

$$2xy^2 = \frac{d}{dx}\left[x^2(y')^2 + (x^2 - \nu^2)y^2\right]$$

Integrating both sides from $0$ to $\alpha$ for a positive $\nu$, we have:

\begin{align*}\int_0^{\alpha}xJ_{\nu}^2(x)\,dx &= \frac{1}{2}\left[x^2(J_{\nu}'(x))^2 + (x^2 - \nu^2)J_{\nu}^2(x)\right]_0^{\alpha}\\&= \frac{\alpha^2}{2}\left[(J_{\nu}'(\alpha))^2 + \left(1 - \frac{\nu^2}{\alpha^2}\right)J_{\nu}^2(\alpha)\right]\\&= \frac{\alpha^2}{2}\left[\left(J_{\nu-1}(\alpha) - \frac{\nu}{\alpha}J_{\nu}(\alpha)\right)^2 + \left(1 - \frac{\nu^2}{\alpha^2}\right)J_{\nu}^2(\alpha)\right]\\&= \frac{\alpha^2}{2}\left[J_{\nu-1}^2(\alpha) - \frac{2\nu}{\alpha}J_{\nu-1}(\alpha)J_{\nu}(\alpha) + J_{\nu}^2(\alpha)\right]\end{align*}

Thus, \begin{align*}I &= \frac{\pi^3}{16}\left[J_{3/2}^2\left(\frac{\pi}{2}\right) - \frac{10}{\pi}J_{3/2}\left(\frac{\pi}{2}\right)J_{5/2}\left(\frac{\pi}{2}\right) + J_{5/2}^2\left(\frac{\pi}{2}\right)\right]\\&= \frac{\pi^3}{16}\left[j_{1}^2\left(\frac{\pi}{2}\right) - \frac{10}{\pi}j_{1}\left(\frac{\pi}{2}\right)j_{2}\left(\frac{\pi}{2}\right) + j_{2}^2\left(\frac{\pi}{2}\right)\right]\\&=\frac{\pi^3}{16}\left[\frac{16}{\pi^4} - \frac{10}{\pi}\frac{4}{\pi^2}\frac{2}{\pi}\left(\frac{12}{\pi^2}-1\right) + \frac{4}{\pi^2}\left(\frac{12}{\pi^2}-1\right)^2\right]\\&= \frac{\pi}{4}-\frac{24}{\pi^3}\end{align*}