Prove $\int_{\mathbb{R^{+}}} \frac{\sin^3 {(\pi x^2)} \cos {(4x^2)}}{x^5} dx=\frac{\pi}{32} (3\pi-4)^2$

Question

How do you arrive at the result $$I=\displaystyle\int_{\mathbb{R^{+}}} \dfrac{\sin^3 {(\pi x^2)} \cos {(4x^2)}}{x^5} dx=\dfrac{\pi}{32} (3\pi-4)^2\ ?$$

Wolfram Alpha agrees numerically.

I tried replacing $\sin p$ by $\dfrac{e^{ip}-e^{-ip}}{2i}$ and similarly for $\cos p$, but in vain.

Answer 1

With the substitution $x^2=t$, $$I=\frac{1}{2}\int_0^{\infty} \frac{\sin^3(\pi t)\cos(4t)}{t^3}\,dt$$ Since, $$\sin(3x)=3\sin x-4\sin^3 x \Rightarrow \sin^3x=\frac{3\sin x-\sin(3x)}{4}$$ $$\Rightarrow \sin^3(\pi t)=\frac{3\sin(\pi t)-\sin(3\pi t)}{4}$$ i.e $$\begin{aligned} I &=\frac{1}{8}\int_0^{\infty} \frac{(3\sin(\pi t)-\sin(3\pi t))\cos(4t)}{t^3}\,dt\\ &=\frac{1}{8}\int_0^{\infty} \frac{3\sin(\pi t)\cos(4t)-\sin(3\pi t)\cos(4t)}{t^3}\,dt\\ &=\frac{1}{16}\int_0^{\infty} \frac{3\sin((\pi+4)t)-3\sin((4-\pi)t)-\sin((3\pi+4)t)-\sin((3\pi-4)t)}{t^3}\,dt \end{aligned}$$ Now use the following: $$\frac{1}{t^3}=\frac{1}{2}\int_0^{\infty} y^2e^{-ty}\,dy$$ and with the result: $$\int_0^{\infty} e^{-ax}\sin(bx)\,dx=\frac{b}{a^2+b^2}$$ we have: $$I=\frac{1}{32}\int_0^{\infty} \left(3(\pi+4)\frac{y^2}{y^2+(\pi+4)^2}-3(4-\pi)\frac{y^2}{y^2+(4-\pi)^2}-(3\pi+4)\frac{y^2}{y^2+(3\pi+4)^2}-(3\pi-4)\frac{y^2}{y^2+(3\pi-4)^2}\right)\,dy$$ Rewrite the terms as follows: $$3(\pi+4)\frac{y^2}{y^2+(\pi+4)^2}=3(\pi+4)-\frac{3(\pi+4)^3}{y^2+(\pi+4)^2}$$ $$3(4-\pi)\frac{y^2}{y^2+(4-\pi)^2}=3(4-\pi)-\frac{3(4-\pi)^3}{y^2+(4-\pi)^2}$$ $$(3\pi+4)\frac{y^2}{y^2+(3\pi+4)^2}=(3\pi+4)-\frac{(3\pi+4)^3}{y^2+(3\pi+4)^2}$$ $$(3\pi-4)\frac{y^2}{y^2+(3\pi-4)^2}=(3\pi-4)-\frac{(3\pi-4)^3}{y^2+(3\pi-4)^2}$$ Hence, $$I=\frac{1}{32}\int_0^{\infty} \left(\frac{(3\pi+4)^3}{y^2+(3\pi+4)^2}+\frac{(3\pi-4)^3}{y^2+(3\pi-4)^2}-\frac{3(\pi+4)^3}{y^2+(\pi+4)^2}+\frac{3(4-\pi)^3}{y^2+(4-\pi)^2}\right)\,dy$$ $$\begin{aligned} \Rightarrow I &=\frac{\pi}{64}\left((3\pi+4)^2+(3\pi-4)^2-3(\pi+4)^2+3(4-\pi)^2\right) \\ &=\frac{\pi}{64}\left(18\pi^2+32-3\cdot 8\pi\right) \\ &= \frac{\pi}{32}\left(9\pi^2+16-24\pi\right) \\ &=\boxed{\dfrac{\pi}{32}\left(3\pi-4\right)^2} \\ \end{aligned}$$

Answer 2

Here I present another way to solve/generalize the problem. It is much easier. Define $$ I_q(p)=\frac{1}{2}\int_0^\infty\frac{\sin^3(pt)\cos(qt)}{t^3}dt,p\ge0,q\ge0. $$ Clearly $I_q(0)=I_q'(0)=I_q''(0)=0$ and \begin{eqnarray} I_q''(p)&=&\frac{3}{8}\int_0^\infty\frac{(3\sin(3pt)-\sin(pt))\cos(qt)}{t}dt\\ &=&\frac{3}{16}\int_0^\infty\frac{3\sin((3p+q)t)+3\sin((3p-q)t)-\sin((p+q)t)-\sin((p-q)t)}{t}dt\\ &=&\left\{\begin{array}{l} 0, \text{ if }p<\frac{q}{3},\\ \frac{9\pi}{32}, \text{ if }p=\frac q3,\\ \frac{9\pi}{16}, \text{ if }\frac{q}{3}<p<q,\\ \frac{15\pi}{32}, \text{ if }p=q,\\ \frac{3\pi}{8}, \text{ if }p>q. \end{array}\right. \end{eqnarray} Here we used $$ \int_0^\infty\frac{\sin(\alpha x)}{x}dx=\text{sgn}(\alpha)\frac{\pi}{2}. $$ Thus $$ I_q'(p)=\left\{\begin{array}{l} 0, \text{ if }p\le\frac{q}{3},\\ \frac{9\pi}{16}(p-\frac{q}{3}), \text{ if }\frac{q}{3}<p\le q,\\ \frac{3p\pi}{8}, \text{ if }p>q, \end{array}\right. $$ and hence $$ I_q(p)=\left\{\begin{array}{l} 0, \text{ if }p\le\frac{q}{3},\\ \frac{\pi}{32}(3p-q)^2, \text{ if }\frac{q}{3}<p\le q,\\ \frac{(3p^2-q^2)\pi}{16}, \text{ if }p>q. \end{array}\right. $$ For $p=\pi,q=4$, since $\frac{q}{3}<p<q$, we have $$ I_4(\pi)=\frac{\pi}{32}(3\pi-4)^2.$$

Answer 3

The following is an approach using contour integration.

$ $

$$ \begin{align} \int_{0}^{\infty} \frac{\sin^{3}(\pi x^{2}) \cos(4x^{2})}{x^{5}} \ dx &= \frac{1}{2}\int_0^{\infty} \frac{\sin^3(\pi t)\cos(4t)}{t^3} \ dt \\ &= \frac{1}{4} \int_{-\infty}^{\infty} \frac{\sin^{3} (\pi t) \cos(4t)}{t^{3}} \ dt \\ &= \frac{1}{4} \text{Re} \int_{-\infty}^{\infty} \frac{\sin^{3} (\pi t) e^{4it}}{t^{3}} \ dt \\ &= \frac{1}{16} \text{Re} \int_{-\infty}^{\infty} \frac{[3 \sin (\pi t) - \sin(3\pi t)]e^{4it}}{t^{3}} \ dt\\ &= \frac{1}{32} \text{Re} \frac{1}{i} \int_{-\infty}^{\infty} \frac{(3e^{\pi i t} - 3e^{ -\pi i t} - e^{3 \pi i t} + e^{-3 \pi i t})e^{4it}}{t^{3}} \ dt \\ &= \frac{1}{32} \text{Re} \frac{1}{i} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} \frac{3e^{(4+\pi) i t} - 3 e^{(4- \pi) i t} - e^{(4+3 \pi) i t} + e^{(4-3 \pi) i t}}{(t-i \epsilon)^{3}} \ dt \\ &= \frac{1}{32} \text{Re} \frac{1}{i} \lim_{\epsilon \to 0^{+}} 2 \pi i \ \text{Res} \left[ \frac{3e^{(4+\pi)iz} - 3 e^{(4- \pi)iz} - e^{(4+ 3 \pi)iz}}{(z- i \epsilon)^{3}},i \epsilon \right] \\ &= \frac{\pi }{16} \text{Re} \lim_{\epsilon \to 0^{+}} \frac{1}{2!} \lim_{z \to i \epsilon} \frac{d^{2}}{dz^{2}} \left( 3e^{(4+ \pi)iz} - 3 e^{(4-\pi)iz} - e^{(4+ 3\pi)iz}\right) \\ &= \frac{\pi}{32} \left(-3( 4+ \pi)^{2} + 3 (4- \pi)^{2} +(4 + 3 \pi)^{2}\right) \\ &= \frac{\pi}{32} \left(3 \pi - 4 \right)^{2} \end{align}$$

EDIT:

Notice that on line 6, $4 - 3 \pi <0$.

So by Jordan's lemma, $\displaystyle \int_{-\infty}^{\infty}\frac{e^{(4- 3\pi)it}}{(z-i \epsilon)^{3}} dt$ would need to be broken off and evaluated separately by closing the contour with the lower half of $|z|=R$ as opposed to the upper half of $|z|=R$.

But since the pole is in the upper half-plane, $\displaystyle \int_{-\infty}^{\infty} \frac{e^{(4- 3\pi)it}}{(t-i \epsilon)^{3}} \ dt $ evaluates to $0$.