# Show $\int_0^\infty\left(1-x\sin\frac 1 x\right)dx = \frac\pi 4$

How to show that $$\int_0^\infty\left(1-x\sin\frac{1}{x}\right)dx=\frac{\pi}{4}$$ ?

• Have you tried a change of variables? – Pedro Tamaroff Sep 4 '12 at 1:09
• $$\int\limits_0^\infty {\left( {1 - \frac{{\sin u}}{u}} \right)\frac{{du}}{{{u^2}}}}$$ seems to be much nicer to deal with, integration by parts, maybe? – Pedro Tamaroff Sep 4 '12 at 1:11
• Taking integration by parts twice to Peter Tamaroff reduces the integral into the famous Dirichlet integral. – Sangchul Lee Sep 4 '12 at 1:51
• It can be solved in many nice ways. (+1) – user 1591719 Sep 5 '12 at 13:38

## 5 Answers

Use $$\int \left(1-x \sin\left(\frac{1}{x}\right)\right) \mathrm{d} x = x - \int \sin\left(\frac{1}{x}\right) \mathrm{d} \frac{x^2}{2} = x - \frac{x^2}{2}\sin\left(\frac{1}{x}\right) - \frac{1}{2} \int \cos\left(\frac{1}{x}\right) \mathrm{d}x$$ Integrating by parts again $\int \cos\left(\frac{1}{x}\right) \mathrm{d}x = x \cos\left(\frac{1}{x}\right) - \int \sin\left(\frac{1}{x}\right) \frac{\mathrm{d}x}{x}$: $$\int \left(1-x \sin\left(\frac{1}{x}\right)\right) \mathrm{d} x = x - \frac{x^2}{2}\sin\left(\frac{1}{x}\right) - \frac{x}{2} \cos\left(\frac{1}{x}\right) + \frac{1}{2} \int \sin\left(\frac{1}{x}\right) \frac{\mathrm{d}x}{x}$$ Thus: $$\begin{eqnarray} \int_0^\infty \left(1-x \sin\left(\frac{1}{x}\right)\right) \mathrm{d} x &=& \left[x - \frac{x^2}{2}\sin\left(\frac{1}{x}\right) - \frac{x}{2} \cos\left(\frac{1}{x}\right)\right]_{0}^{\infty} + \frac{1}{2} \int_0^\infty\sin\left(\frac{1}{x}\right) \frac{\mathrm{d}x}{x} = \\ &=& 0 + \frac{1}{2} \int_0^\infty \frac{\sin{u}}{u} \mathrm{d} u = \frac{\pi}{4} \end{eqnarray}$$ where the last integral is the Dirichlet integral.

Sasha's answer concisely gets the answer in terms of the Dirichlet integral, so I will evaluate this integral in the same way that the Dirichlet integral is evaluated with contour integration.

First, change variables to $z=1/x$: $$\int_0^\infty\left(1-x\sin\left(\frac1x\right)\right)\,\mathrm{d}x =\int_0^\infty\frac{z-\sin(z)}{z^3}\,\mathrm{d}z\tag{1}$$ Since the integrand on the right side of $(1)$ is even, entire, and vanishes as $t\to\infty$ within $1$ of the real axis, we can use symmetry to deduce that the integral is $\frac12$ the integral over the entire line and then shift the path of integration by $-i$: $$\int_0^\infty\frac{z-\sin(z)}{z^3}\,\mathrm{d}z =\frac12\int_{-\infty-i}^{\infty-i}\frac{z-\sin(z)}{z^3}\,\mathrm{d}z\tag{2}$$ Consider the contours $\gamma^+$ and $\gamma^-$ below. Both pass a distance $1$ below the real axis and then circle back along circles of arbitrarily large radius.

$\hspace{4.4cm}$ Next, write $\sin(z)=\frac1{2i}\left(e^{iz}-e^{-iz}\right)$ and split the integral as follows $$\frac12\int_{-\infty-i}^{\infty-i}\frac{z-\sin(z)}{z^3}\,\mathrm{d}z =\frac12\int_{\gamma^-}\left(\frac1{z^2}+\frac{e^{-iz}}{2iz^3}\right)\,\mathrm{d}z -\frac12\int_{\gamma^+}\frac{e^{iz}}{2iz^3}\,\mathrm{d}z\tag{3}$$ $\gamma^-$ contains no singularities so the integral around $\gamma^-$ is $0$. The integral around $\gamma^+$ is $\color{#00A000}{2\pi i}$ times $\color{#00A000}{-\dfrac{1}{4i}}$ times the residue of $\color{#C00000}{\dfrac{e^{iz}}{z^3}}$ at $\color{#C00000}{z=0}$; that is, $\color{#00A000}{-\dfrac\pi2}$ times the coefficient of $\color{#C00000}{\dfrac1z}$ in $$\frac{1+iz\color{#C00000}{-z^2/2}-iz^3/6+\dots}{\color{#C00000}{z^3}}\tag{4}$$ Thus, the integral around $\gamma^+$ is $\color{#00A000}{\left(-\dfrac\pi2\right)}\color{#C00000}{\left(-\dfrac12\right)}=\dfrac\pi4$. Therefore, combining $(1)$, $(2)$, and $(3)$ yields $$\int_0^\infty\left(1-x\sin\left(\frac1x\right)\right)\,\mathrm{d}x=\frac\pi4\tag{5}$$ As complicated as that may look at first glance, with a bit of practice, it is easy enough to do in your head.

Let's start out with the variable change $\displaystyle x=\frac{1}{u}$ and then turn the integral into a double integral: $$\int_{0}^{\infty} {\left( {1 - \frac{\sin u}{u}} \right)\frac{1}{u^2}} \ du=$$ $$\int_{0}^{\infty}\left(\int_{0}^{1} 1 - \cos (u a) \ da \right)\frac{1}{u^2} \ du=$$ By changing the integration order we get $$\int_{0}^{1}\left(\int_{0}^{\infty} \frac{1 - \cos (a u)}{u^2} \ du \right)\ \ da=\int_{0}^{1} a \frac{\pi}{2} \ da=\frac{\pi}{4}.$$

Note that by using a simple integration by parts at $\displaystyle \int_{0}^{\infty} \frac{1 - \cos (a u)}{u^2} \ du$ we immediately get $\displaystyle a\int_{0}^{\infty} \frac{\sin(au)}{u} \ du = a\int_{0}^{\infty} \frac{\sin(u)}{u}\ du$ that is $\displaystyle a\frac{\pi}{2}$. The last integral is the famous Dirichlet integral.
Hence the result follows and the proof is complete.

Q.E.D. (Chris)

Another way is to use Laplace transform to evaluate this improper integral. In fact, let $f(x)=x-\sin x$ and then $F(s)=\frac{1}{s^2}-\frac{1}{s^2+1}$. Thus \begin{eqnarray*} \mathcal{L}\big\{\frac{f(x)}{x}\big\}&=&\int_s^\infty F(s)ds=-\frac{\pi}{2}+\frac{1}{s}+\arctan s, \\ \mathcal{L}\big\{\frac{f(x)}{x^2}\big\}&=&\int_s^\infty(-\frac{\pi}{2}+\frac{1}{u}+\arctan u)du\\ &=&-1+s(\frac{\pi}{2}-\arctan s)+\ln\frac{\sqrt{s^2+1}}{s}, \\ \mathcal{L}\big\{\frac{f(x)}{x^3}\big\}&=&\int_s^\infty\left(-1+u(\frac{\pi}{2}-\arctan u)-\ln\frac{\sqrt{u^2+1}}{u}\right)du\\ &=&\frac{s}{2}-\frac{\pi s^2}{4}-\frac{1}{2}\arctan s+\frac{1}{2}s^2\arctan s+s\ln s-\frac{1}{2}s\ln(s^2+1). \end{eqnarray*} From this, one can obtain \begin{eqnarray*} &&\int_0^\infty (1-x\sin\frac{1}{x})dx=\int_0^\infty\frac{x-\sin x}{x^3}dx=\lim_{s\to 0^+}\mathcal{L}\big\{\frac{f(x)}{x^3}\big\}\\ &=&\lim_{s\to 0^+}(\frac{\pi}{4}+\frac{s}{2}-\frac{\pi s^2}{4}-\frac{1}{2}\arctan s+\frac{1}{2}s^2\arctan s+s\ln s-\frac{1}{2}s\ln(s^2+1))\\ &=&\frac{\pi}{4}. \end{eqnarray*}

An idea with complex integration. First, we change variables:

$$x=\frac{1}{u}\Longrightarrow dx=-\frac{du}{u^2}\Longrightarrow \int_0^\infty\left(1-x\sin\frac{1}{x}\right)dx=-\int_\infty^0\left(1-\frac{1}{u}\sin u\right)\frac{du}{u^2}=$$

$$=\int_0^\infty\frac{x-\sin x}{x^3}dx$$

Define (with circular paths always in the positive anticlockwise direction)

$$f(z)=\frac{iz-e^{iz}}{z^3}\,\,,\,\gamma_r:=\{re^{it}\;|\;r>0\,,\,0\leq t\leq \pi\}$$

$$\Gamma:=[-R,-\epsilon]\cup\left(-\gamma_\epsilon\right)\cup[\epsilon,R]+\gamma_R$$

We also calculate the residue of the function at $\,z=0\,$ by means of power series:

$$f(z)=\frac{iz-e^{iz}}{z^3}=\frac{1}{z^3}\left[iz-\left(1+iz-\frac{z^2}{2}-\frac{iz^3}{6}+...\right)\right]=\frac{1}{z^3}\left(-1+\frac{z^2}{2}+..\right)=$$

$$=-\frac{1}{z^3}+\frac{1}{2z}+...\Longrightarrow\operatorname{Res}_{z=0}(f)=\frac{1}{2}$$ We use the above, and the lemma and its corollary in the answer here , to get:

$$\lim_{\epsilon\to 0}\int_{\gamma_\epsilon}f(z)\,dz=\frac{\pi i}{2}$$

We also have

$$\left|\int_{\gamma_R}f(z)\,dz\right|\leq\max_{z\in\gamma_R}\frac{|iz-e^{iz}|}{|z^3|}R\pi\leq \max_{z=Re^{it}}\frac{R+e^{-R\sin t}}{R^2}\pi\xrightarrow [R\to\infty]{}0$$

(Note: $\sin t\geq 0\,$ when $\,t\in [0,\pi]\,$)

Thus, by Cauchy's Formula and the above, and since $\,f(z)\,$ is analytic in the domain enclosed by $\,\Gamma\,$:

$$0=\int_\Gamma f(z)\,dz=\int_{-R}^\epsilon f(x)\,dx-\int_{\gamma_\epsilon}f(z)\,dz+\int_\epsilon^R f(x)\,dx+\int_{\gamma_R}f(z)\,dz$$

And passing now to the limits when $\,\epsilon\to 0\,\,,\,\,R\to \infty\,$ , we get

$$0=\int_{-\infty}^\infty\frac{ix-e^{ix}}{x^3}-\frac{\pi i}{2}\stackrel{\text{equalling imaginary parts}}\Longrightarrow \int_{-\infty}^\infty\frac{x-\sin x}{x^3}=\frac{\pi}{2}$$

And since the integrand is an even function, we finally get

$$\int_0^\infty\left(1-x\sin\frac{1}{x}\right)dx=\int_0^\infty\frac{x-\sin x}{x^3}dx=\frac{\pi}{4}$$