How do I show that $\int_0^\infty \frac{\sin(ax) \sin(bx)}{x^{2}} \, \mathrm dx = \pi \min(a,b)/2$

Solution 1:

In several steps:

• Trigonometric relation $$\left( \frac{\sin ax}{x}\right)\left( \frac{\sin bx}{x}\right)=\frac{1-\cos(a+b)x}{2x^2}-\frac{1-\cos(a-b)x}{2x^2}$$
• Dirichlet integral $$\int_0^\infty \frac{\sin \alpha t}{t}dt=\frac{\pi}{2}\mathrm{sgn}(\alpha)$$
• Integration by parts $$\int_0^\infty\frac{1-\cos(\alpha)t}{t^2}=\alpha\int_0^\infty \frac{\sin \alpha t}{t}dt=\frac{\pi}{2}|\alpha|$$
• Combine $$\int_0^\infty \left( \frac{\sin ax}{x}\right)\left( \frac{\sin bx}{x}\right) \mathrm{d}x=\frac{\pi}{4}(|a+b|-|a-b|)=\frac{\pi}{2}\min(a,b)$$

Solution 2:

One way is to to this by residues. Another way to integrate once by parts to get $$I=\int_0^{\infty}\frac{b\sin ax\cos bx+a\cos ax \sin bx}{x}dx,$$ then to use the formula $2\sin \alpha\cos\beta=\sin(\alpha+\beta)+\sin(\alpha-\beta)$ and the mentioned integral (note that your formula needs to be corrected on the left and on the right) $$\displaystyle \int_0^{\infty}\frac{\sin xy}{x}\,dx=\frac{\pi}{2}\mathrm{sgn}(y).$$ This gives \begin{align}I&=\frac{\pi}{4}\Bigl[b\,\mathrm{sign}(a+b)+b\,\mathrm{sign}(a-b)+a\,\mathrm{sign}(a+b)+a\,\mathrm{sign}(b-a)\Bigr]=\\ &=\frac{\pi}{4}\left(|a+b|-|a-b|\right), \end{align} For $a,b>0$ the last expression is obviously equal to $\pi\min\{a,b\}/2$.

Solution 3:

A very easy way to see this is to use Parseval's theorem for Fourier transforms. In general, Parseval's theorem states that, for two functions $f$ and $g$, each having respective FTs $\hat{f}$ and $\hat{g}$, related by

$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \, f(x) \, e^{i k x}$$

etc., then

$$\int_{-\infty}^{\infty} dx \, f(x) \bar{g}(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \, \hat{f}(k) \bar{\hat{g}}(k)$$

The FT of $\sin{(a x)}/x$ is given by

$$\int_{-\infty}^{\infty} dx \, \frac{\sin{a x}}{x} e^{i k x} = \begin{cases} \pi & |k| \lt a \\ 0 & |k| \gt a\end{cases}$$

Similarly,

$$\int_{-\infty}^{\infty} dx \, \frac{\sin{b x}}{x} e^{i k x} = \begin{cases} \pi & |k| \lt b \\ 0 & |k| \gt b\end{cases}$$

By Parseval, we take the integral of the product of the transforms, which is the product of two rectangles. The product is clearly nonzero over the smaller of the two widths, i.e. $2 \min\{a,b\}$. Thus,

$$\int_{-\infty}^{\infty} dx \, \frac{\sin{a x}}{x} \, \frac{\sin{b x}}{x} = \frac{\pi^2}{2 \pi} 2 \min\{a,b\} $$

Therefore

$$\int_{0}^{\infty} dx \, \frac{\sin{a x}}{x} \, \frac{\sin{b x}}{x} = \frac{\pi}{2} \min\{a,b\}$$

Solution 4:

Let $b_{k} >0$ and $a \ge \sum_{k=1}^{n} b_{k}$.

Generalizing the answer HERE, we can use contour integration to quickly show that

$$\int_{0}^{\infty} \frac{\sin(ax)}{x} \prod_{k=1}^{n} \frac{\sin \left( b_{k}x \right)}{x} \, dx= \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin(ax)}{x} \prod_{k=1}^{n} \frac{\sin \left( b_{k}x \right)}{x} \, dx = \frac{\pi}{2} \prod_{k=1}^{n}b_{k}. \tag{1}$$

Under the conditions stated above, the function $$e^{iaz} \prod_{k=1}^{n} \sin \left( b_{k}x \right) = e^{iaz} \prod_{k=1}^{n} \frac{e^{ib_{k}z}-e^{-ib_{k}z} }{2i}$$ is bounded in the upper half-plane.

So by integrating the function $$f(z) = \frac{e^{iaz}}{z} \prod_{k=1}^{n} \frac{\sin \left( b_{k}z \right)}{z} $$ around an indented contour that consists of the real axis and the semiciricle above it, we get (in the limit), $$\text{PV} \int_{-\infty}^{\infty} \frac{e^{iax}}{x} \prod_{k=1}^{n} \frac{\sin \left( b_{k}x \right)}{x} \, dx - \pi i \, \text{Res} [f(z), 0] = 0,$$ where $$\text{Res}[f(z) ,0] = \lim_{z \to 0}e^{iaz} \prod_{k=1}^{n}\frac{\sin \left( b_{k}z \right)}{z} =1\times \prod_{k=1}^{n} b_{k}. $$

Taking the imaginary parts of both sides of equation leads to the result.

Solution 5:

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It's sufficient to consider the case $\ds{\bbox[10px,#ffe,border:1px dotted navy] {\ds{a, b \in \mathbb{R}_{\ >\ 0}}}}$.

\begin{align} \mbox{Note that}\ &\int_{0}^{\infty}{\sin\pars{ax} \over x}\,{\sin\pars{bx} \over x}\,\dd x = {1 \over 2}\,b\int_{-\infty}^{\infty}{\sin\pars{ax} \over ax} \,{\sin\pars{\bracks{b/a}ax} \over \pars{b/a}ax}\,a\,\dd x \\[5mm] = &\ {1 \over 2}\,b\int_{-\infty}^{\infty}{\sin\pars{x} \over x} \,{\sin\pars{x/\mu} \over x/\mu}\,\dd x \,,\qquad\mu \equiv {a \over b} > 0 \label{1}\tag{1} \end{align}