# Hints with this Integration Problem

$$\int_{0}^{\infty}\frac{(x \cos x-\sin x)\cos (x/2)}{x^3}\mathrm d x$$

I tried solving it by substitution, and I did come close to the answer. But that was really long and tedious, I don't even know whether or not what I did was even correct.

Can anyone suggest me some other way? Hints, anything?

Cuz I can't think of anything other than substitution, thoughts?

• perhaps contour integral will help you – haqnatural Jul 13 '18 at 18:42
• [ Wolfy]( wolframalpha.com/input/?i=integrate+(x*cos(x)-sin(x))%2Fx%5E3*cos(x%2F2) ) $\text{Si}(0)=0$ and $\text{Si}(0)=\frac{\pi}{2}$ – FDP Jul 13 '18 at 19:13
• @FDP tbh integral calculator gives an answer too, but I'm not after the answer, I'm here to learn the method to evaluate this. – William Jul 13 '18 at 19:16
• An anti-derivative does exist. If you don't trust Wolfy, derive its result. It's enough to terminate your computation. Anyway the result gives a path to follow. – FDP Jul 13 '18 at 19:20
• Probably integration by parts is required – FDP Jul 13 '18 at 19:25

This may be done by recognizing that

$$\frac{x \cos{x}-\sin{x}}{x^2} = \frac{d}{dx} \frac{\sin{x}}{x} = \frac{d}{dx} \int_0^1 du \, \cos{x u} = -\int_0^1 du \, u \sin{x u}$$

Thus the integral is then equal to

$$-\int_0^{\infty} dx \frac{\cos{(x/2)}}{x} \, \int_0^1 du \, u \sin{x u}$$

We can reverse the order of integration to get

$$-\int_0^1 du \, u \, \int_0^{\infty} dx \frac{\sin{x u}}{x} \cos{(x/2)}$$

which looks like a Fourier transform:

$$\int_0^{\infty} dx \frac{\sin{x u}}{x} \cos{(x/2)} = \frac12 \int_{-\infty}^{\infty} dx \, \frac{\sin{x u}}{x} e^{i k x}$$

where $k = \frac1{2}$. Thus, the integral is simply $\pi$ when $|u|>1/2$ and zero elsewhere. Thus the integral is equal to

$$-\frac{\pi}{2} \int_{1/2}^1 du \, u = -\frac{3 \pi}{16}$$

• Are you sure it's $\int_{\infty}^{\infty}$ and not $\int_{- \infty}^{\infty}$ at the end? – William Jul 13 '18 at 20:10
• @William: thanks for catching the typo (and for posing the question that way). – Ron Gordon Jul 13 '18 at 20:11

\begin{align}J&=\int_{0}^{\infty}\frac{(x \cos x-\sin x)\cos (x/2)}{x^3}\mathrm d x\end{align}

Perform integration by parts,

\begin{align}J&=\left[-\frac{1}{2x^2}(x \cos x-\sin x)\cos (x/2)\right]_{0}^{\infty}+\\ &\int_0^{\infty}\frac{-\frac{1}{2}\sin\left( \frac{1}{2}x\right)(x \cos x-\sin x)-x\sin x\cos\left( \frac{1}{2}x\right)}{2x^2}dx\\ &=\int_0^{\infty}\frac{-\frac{1}{2}\sin\left( \frac{1}{2}x\right)(x \cos x-\sin x)-x\sin x\cos\left( \frac{1}{2}x\right)}{2x^2}\,dx\\ &=-\frac{1}{4}\int_0^{\infty}\frac{\sin\left( \frac{1}{2}x\right)\cos x}{x}\,dx+\frac{1}{4}\int_0^{\infty}\frac{\sin\left( \frac{1}{2}x\right)\sin x}{x^2}\,dx-\frac{1}{2}\int_0^{\infty}\frac{\cos\left( \frac{1}{2}x\right)\sin x}{x}\,dx\\ &=-\frac{1}{4}\int_0^{\infty}\frac{\sin\left( \frac{1}{2}x\right)\cos x}{x}\,dx-\frac{1}{2}\int_0^{\infty}\frac{\cos\left( \frac{1}{2}x\right)\sin x}{x}\,dx+\left[-\frac{\sin\left( \frac{1}{2}x\right)\sin x}{4x}\right]_0^{\infty}+\\ &\frac{1}{4}\int_0^{\infty}\frac{\sin\left( \frac{1}{2}x\right)\cos x}{x}\,dx+\frac{1}{8}\int_0^{\infty}\frac{\cos\left( \frac{1}{2}x\right)\sin x}{x}\,dx\\ &=-\frac{3}{8}\int_0^{\infty}\frac{\cos\left( \frac{1}{2}x\right)\sin x}{x}\,dx\\ &=-\frac{3}{16}\int_0^{\infty}\frac{\sin\left( \frac{3}{2}x\right)+\sin\left( \frac{1}{2}x\right)}{x}\,dx\\ &=-\frac{3}{16}\int_0^{\infty}\frac{\sin\left( \frac{3}{2}x\right)}{x}\,dx-\frac{3}{16}\int_0^{\infty}\frac{\sin\left( \frac{1}{2}x\right)}{x}\,dx \end{align}

In the first intégral perform the change of variable $y=\dfrac{3}{2}x$,

in the second intégral perform the change of variable $y=\dfrac{1}{2}x$,

\begin{align}J&=-\frac{3}{16}\int_0^{\infty}\frac{\sin x }{x}\,dx-\frac{3}{16}\int_0^{\infty}\frac{\sin x }{x}\,dx\\ &=-\frac{3}{8}\int_0^{\infty}\frac{\sin x }{x}\,dx\\ \end{align}

But,

It's well-known that,

\begin{align}\int_0^{\infty}\frac{\sin x }{x}\,dx=\frac{\pi}{2}\end{align}

Therefore,

\begin{align}J&=-\frac{3}{8}\times \frac{\pi}{2}\\ &=\boxed{-\frac{3}{16}\pi} \end{align}