What is the use of Dirichlet Integral?

Question

How can I find the value of

$$\large\int_0^\infty\left(\dfrac{\sin x}x\right)^5dx$$ using Contour Integrals? I attempted it using Integration by Parts and got the an got the answer. I have studied the basics of contour integration...

Further, how can I generalise it to the following? $$\large\int_0^\infty\left(\dfrac{\sin x}x\right)^ndx$$

Answer 1

HINT:

We provide herein only an outline of the approach to evaluating the integral of interest using contour integration.

Step 1: First note that $\frac{\sin z}{z}$ is an even function. Therefore,

$$\int_0^\infty\frac{\sin^5 x}{x^5}\,dx=\frac12 \int_{-\infty}^\infty\frac{\sin^5 x}{x^5}\,dx$$

Step 2: Use Euler's Identity along with the Binomial Theorem to write

$$\sin^5 x=\frac{e^{i5z}}{32i}\,\sum_{k=0}^5\binom{5}{k}(-1)^k\,e^{-i2kz}$$

Step 3: Note for $n>0$ ($n<0$) we have from the residue theorem

$$\begin{align} \oint_C \frac{e^{inz}}{z^5}\,dz&=\text{sgn}(n)\,2\pi i \text{Res}\left(\frac{e^{inz}}{z^5},z=0\right)\\\\ &=2\pi i \text{sgn}(n) \left(\frac{n^4}{4!}\right) \end{align}$$

where $C$ is a contour consisting of (i) the real-line segments from $(-R,0)$ to $(-\epsilon,0)$ and from $(\epsilon,0)$ to $(R,0)$, (ii) a semi-circle in the lower-half (upper-half) plane, centered at the origin with radius $\epsilon$, and (iii) a semi-circle in the upper-half (lower-half) plane, centered at the origin with radius $R$.

Then, note finally that

$$\lim_{R\to \infty}\lim_{\epsilon \to 0}\oint_C\frac{e^{inz}}{z^5}\,dz=\text{PV}\left(\int_{-\infty}^\infty\frac{e^{inx}}{x^5}\,dx\right)+\text{sgn}(n)\pi i \frac{n^4}{4!}$$

The rest should be straightforward.

Answer 2

One of the method is to use Integration By Parts...

• Dirichlet Integral: $$\displaystyle\int_0^\infty \dfrac{\sin x}x \,dx= \dfrac{\pi}2$$

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• $\displaystyle \int_0^\infty \dfrac{\sin^3x}xdx=\dfrac\pi4$ :

We know $\sin^3x = \dfrac{3\sin x -\sin3x}4$ $$\begin{align*} \int_0^\infty \dfrac{\sin^3x}xdx &=\dfrac14 \int_0^\infty\dfrac{3\sin x}xdx-\int_0^\infty\dfrac{\sin3x}xdx\\&=\dfrac{3\pi}8-\int_0^\infty \dfrac{\sin y}ydy\\&=\dfrac\pi4\end{align*}$$

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• $\displaystyle \int_0^\infty \dfrac{\sin^5x}xdx=\dfrac{3\pi}{16}$

We know $\sin^5x=\dfrac{\sin 5x +20\sin^3x-5\sin x}{16}$ $$\begin{align*}\int_0^\infty\dfrac{\sin^5x}xdx&=\dfrac1{16}\left[\int_0^\infty \dfrac{\sin5x}xdx+20\int_0^\infty\dfrac{\sin^3x}xdx - 5\int_0^\infty\dfrac{\sin x}xdx\right]\\&=\dfrac1{16}\left[\dfrac \pi 2 + 5\pi - \dfrac{5\pi}2\right]\\&=\dfrac{3\pi}{16}\end{align*}$$

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• Since the integrations by parts is from $0\to\infty$ we can ignore the term $uv$ in $$\int u\,dv=uv-\int v\,du$$

So, we can use the following here(not everywhere). You can use this here because $$\lim_{x\to\infty}\int v(x)dx = \lim_{x\to\infty}\dfrac 1{x^n}=0 \text{, for some }n > 0 $$ and $$u(0)=\sin 0=0$$ So, $$\int_0^\infty u(x)v(x)dx = -\int_0^\infty \left(u'(x)\int v(x)dx\right)dx$$

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• $\displaystyle\int_0^\infty \dfrac{sin^3x}{x^3}dx = \dfrac{3\pi}8$

$$\displaystyle\begin{align*}\int_0^\infty \dfrac{\sin^3x}{x^3}dx&=\dfrac32\int_0^\infty\dfrac{\sin^2x\cos x}{x^2}dx\\&=\dfrac32\int_0^\infty\dfrac{2\sin x\cos^2x-sin^3x}xdx\\&=\dfrac32\left[2\int_0^\infty\dfrac{\sin x}xdx-3\int_0^\infty\dfrac{\sin^3x}xdx\right]\\&=\dfrac32\left(\pi-\dfrac{3\pi}4\right)\\&=\dfrac{3\pi}8\end{align*}$$

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Let's do Parts! $$\begin{align*} \int_0^\infty \left( \frac{\sin x}x \right)^5 \, dx &=\dfrac 54 \int_0^\infty \dfrac{\sin^4x\cos x}{x^4}dx\\&=\dfrac5{12} \int_0^\infty \dfrac{4\sin^3x\cos^2x-\sin^5x}{x^3}dx\end{align*} $$ Using $\cos^2x=1-\sin^2x$$$\begin{align*} \int_0^\infty \left( \frac{\sin x}x \right)^5 \, dx&=\dfrac5{12}\left[ \int_0^\infty \dfrac{4\sin^3x}{x^3}dx- \int_0^\infty \dfrac{5\sin^5}{x^3}dx\right]\end{align*} $$ Using $\displaystyle\int_0^\infty \dfrac{sin^3x}{x^3}dx = \dfrac{3\pi}8$, $$\begin{align*} \int_0^\infty \left( \frac{\sin x}x \right)^5 \, dx&=\dfrac{5\pi}8-\dfrac{25}{12} \int_0^\infty \dfrac{\sin^5x}{x^3}dx\end{align*} $$

Repeating the Integration By Parts again, $$\begin{align*} \int_0^\infty \left( \frac{\sin x}x \right)^5 \, dx&=\dfrac{5\pi}{8}-\dfrac{125}{24}\left[ \int_0^\infty \dfrac{4\sin^3x}xdx- \int_0^\infty \dfrac{5\sin^5x}xdx\right]\end{align*} $$

Using $\displaystyle \int_0^\infty \dfrac{\sin^3x}xdx=\dfrac\pi4$ and $\displaystyle \int_0^\infty \dfrac{\sin^5x}xdx=\dfrac{3\pi}{16}$

So, $$ \int_0^\infty \dfrac{4\sin^3x}xdx- \int_0^\infty \dfrac{5\sin^5x}xdx =\pi-\dfrac{15\pi}{16}= \dfrac{\pi}{16}$$

$$\begin{align*} \Rightarrow\int_0^\infty \left( \frac{\sin x}x \right)^5 \, dx&=\dfrac{5\pi}8-\dfrac{125\pi}{24\times16}\\&=\dfrac{115\pi}{384}\end{align*} $$

$$\displaystyle\therefore\int_0^\infty \left( \frac{\sin x}x \right)^5 dx =\dfrac{115\pi}{384}$$