6
$\begingroup$

Calculate $\displaystyle\int_0^\infty\dfrac{\sin(x)\log(x)}{x}\mathrm dx$.

I tried to expand $\sin(x)$ at zero, or use SI(SinIntegral) function, but it did not work. Besides, I searched the question on math.stackexchange, nothing found.

Mathematica tells me the answer is $-\dfrac{\gamma\pi}{2}$, I have no idea how to get it.

Thanks for your help!

$\endgroup$
16
$\begingroup$

Notice that

\begin{align*} \int_{0}^{\infty} \frac{\sin x}{x^s} \, dx &= \frac{1}{\Gamma(s)} \int_{0}^{\infty} \left( \int_{0}^{\infty} t^{s-1}e^{-xt} \, dt \right) \sin x \, dx \\ &= \frac{1}{\Gamma(s)} \int_{0}^{\infty} \left( \int_{0}^{\infty} e^{-tx} \sin x \, dx \right) t^{s-1} \, dt \\ &= \frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{t^{s-1}}{1+t^2} \, dt \\ &= \frac{1}{2\Gamma(s)} \beta\left(\frac{s}{2}, 1-\frac{s}{2}\right) \\ &= \frac{\pi}{2\Gamma(s)\sin\left(\frac{\pi s}{2}\right)}. \end{align*}

(This heuristic computation is valid line-by-line on the strip $1 < \Re(s) <2$ in view of Fubini's theorem, and then extends to the larger strip $0 < \Re(s) < 2$ by analytic continuation.) Differentiating both sides, we get

$$ \int_{0}^{\infty} \frac{\sin x \log x}{x^s} \, dx \stackrel{(*)}{=} -\frac{d}{ds} \frac{\pi}{2\Gamma(s)\sin\left(\frac{\pi s}{2}\right)} = \frac{\pi}{2\Gamma(s)\sin\left(\frac{\pi s}{2}\right)} \left( \psi(s) + \frac{\pi}{2}\cot\left(\frac{\pi s}{2}\right) \right). $$

Now the answer follows by plugging $s = 1$:

$$ \int_{0}^{\infty} \frac{\sin x \log x}{x} \, dx = -\frac{\gamma \pi}{2}. $$


Justification of $\text{(*)}$. Let us prove that

$$F(s) := \int_{0}^{\infty} \frac{\sin x}{x^s} \, dx \tag{1}$$

is analytic on the open strip $S = \{ s \in \Bbb{C} : 0 < \Re(s) < 2 \}$ and its derivative can be computed by the Leibniz's integral rule. A major issue of $\text{(1)}$ is that the defining integral converges only conditionally for $0 < \Re(s) \leq 1$ and thus raises some technical difficulties. In order to circumvent this, we improve the speed of convergence using integration by parts:

$$ F(s) = \underbrace{\left[ \frac{1-\cos x}{x^s} \right]_{0}^{\infty}}_{=0} + s \int_{0}^{\infty} \frac{1-\cos x}{x^{s+1}} \, dx. $$

Notice that the resulting integral is absolutely convergent on $S$.

Now we claim that $g(s) := F(s)/s$ is differentiable and its derivative can be computed by the Leibniz's integral rule. Let $\epsilon$ be such that $\bar{B}(s, \epsilon) \subset S$. Then whenever $0 < |h| < \epsilon$,

\begin{align*} &\frac{g(s+h) - g(s)}{h} - \int_{0}^{\infty} \frac{(1-\cos x)(-\log x)}{x^{1+s}} \, dx \\ &\hspace{9em} = \int_{0}^{\infty} \frac{1-\cos x}{x^{1+s}} \left( \frac{x^{-h} - 1}{h} + \log x \right) \, dx. \end{align*}

Notice that the integrand is dominated by

$$ \left| \frac{1-\cos x}{x^{1+s}} \left( \frac{x^{-h} - 1}{h} + \log x \right) \right| \leq \frac{1-\cos x}{x^{1+\Re(s)}} (\max\{ x^{\epsilon}, x^{-\epsilon} \} + 1) |\log x| $$

This dominating function is integrable. Hence by the dominated convergence theorem, as $h \to 0$, we have

$$ \lim_{h \to 0} \frac{g(s+h) - g(s)}{h} = \int_{0}^{\infty} \frac{(1-\cos x)(-\log x)}{x^{1+s}} \, dx. $$

Plugging this back, we know that $F(s)$ is differentiable with

$$ F'(s) = \int_{0}^{\infty} \frac{1-\cos x}{x^{s+1}} \, dx + s \int_{0}^{\infty} \frac{(1-\cos x)(-\log x)}{x^{1+s}} \, dx. $$

Finally, performing integration by part to the latter integral yields the desired conclusion:

\begin{align*} &s \int_{0}^{\infty} \frac{(1-\cos x)(-\log x)}{x^{1+s}} \, dx \\ &\hspace{5em} = \underbrace{\left[ \frac{(1-\cos x)\log x}{x^s} \right]_{0}^{\infty}}_{=0} - \int_{0}^{\infty} \left( \frac{\sin x \log x}{x^s} + \frac{1-\cos x}{x^{1+s}} \right) \, dx \end{align*}

$\endgroup$
3
  • $\begingroup$ @JackD'Aurizio, Thank you! $\endgroup$ – Sangchul Lee Sep 26 '16 at 13:35
  • $\begingroup$ Nothing to thank me for, it is a really well-written answer, credits to you. $\endgroup$ – Jack D'Aurizio Sep 26 '16 at 13:38
  • $\begingroup$ +1. Not too many people care to write a justification $^{*}$ as you did it ( included me !!! ). $\endgroup$ – Felix Marin Oct 2 '16 at 8:37
5
$\begingroup$

$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\color{#f00}{\int_{0}^{\infty}{\sin\pars{x}\ln\pars{x} \over x}\,\dd x} = \int_{0}^{\infty}\sin\pars{x}\ln\pars{x}\int_{0}^{\infty}\expo{-xt}\,\dd t\,\dd x \\[5mm] = &\ \int_{0}^{\infty}\ \overbrace{\Im\int_{0}^{\infty}\ln\pars{x}\expo{-\pars{t - \ic}x}\,\dd x} ^{\ds{\equiv\ \mc{J}}}\ \,\dd t\label{1}\tag{1} \end{align}

Hereafter the $\ds{\ln}$-branch-cut runs along $\ds{\left(-\infty,0\right]}$ with $\ds{\ln\pars{z}\ \,\mrm{arg}}$ given by $\ds{-\pi < \mrm{arg}\pars{z} < \pi}$.


\begin{align} \mc{J} & \equiv \Im\int_{0}^{\infty}\ln\pars{x}\expo{-\pars{t - \ic}x}\,\dd x = \Im\bracks{{1 \over t - \ic}\int_{0}^{\pars{t - \ic}\infty} \ln\pars{x \over t - \ic}\expo{-x}\,\dd x} \\[5mm] & = -\,\Im\braces{{t + \ic \over t^{2} + 1}\int_{\infty}^{0}\bracks{\ln\pars{x \over \root{t^{2} + 1}} + \arctan\pars{1 \over t}\ic}\expo{-x}\,\dd x} \\[5mm] & = \Im\braces{{t + \ic \over t^{2} + 1}\bracks{% \int_{0}^{\infty}\ln\pars{x}\expo{-x}\,\dd x - {1 \over 2}\ln\pars{t^{2} + 1}+ \arctan\pars{1 \over t}\ic}} \end{align}

Note that
$\ds{\int_{0}^{\infty}\ln\pars{x}\expo{-x}\,\dd x = \left.\partiald{}{\mu}\int_{0}^{\infty}x^{\mu}\expo{-x}\,\dd x\, \right\vert_{\ \mu\ =\ 0} = \Gamma\,'\pars{1}\ =\ \overbrace{\Gamma\pars{1}}^{\ds{=\ 1}}\ \overbrace{\Psi\pars{1}}^{\ds{-\gamma}}\ =\ -\gamma\quad}$ where $\ds{\Gamma}$

and $\ds{\Psi}$ are the Gamma and Digamma Functions, respectively. Then,

\begin{align} \mc{J} & \equiv \bbox[8px,#ffe,border:0.1em groove navy]{% \Im\int_{0}^{\infty}\ln\pars{x}\expo{-\pars{t - \ic}x}\,\dd x} = -\,{\gamma \over t^{2} + 1} - {1 \over 2}\,{\ln\pars{t^{2} + 1} \over t^{2} + 1} + {t\arctan\pars{1/t} \over t^{2} + 1} \\[5mm] & = \bbox[8px,#ffe,border:0.1em groove navy]{-\,{\gamma \over t^{2} + 1} + \totald{}{t}\bracks{{1 \over 2}\,\arctan\pars{1 \over t}\ln\pars{t^{2} + 1}}} \label{2}\tag{2} \end{align}

Note that$\ds{{1 \over 2}\,\arctan\pars{1 \over t}\ln\pars{t^{2} + 1}}$ vanishes out when $\ds{t \to \infty}$ and $\ds{t \to 0^{+}}$.

By replacing \eqref{2} in \eqref{1}: \begin{align} &\color{#f00}{\int_{0}^{\infty}{\sin\pars{x}\ln\pars{x} \over x}\,\dd x} = -\gamma\int_{0}^{\infty}{\dd t \over t^{2} + 1} = \color{#f00}{-\,{1 \over 2}\,\gamma\pi} \end{align}

$\endgroup$
1
  • $\begingroup$ (+1) That is how I just did this. Then I looked on MSE to see if it's already been discusses and found this page. Well done my friend! $\endgroup$ – Mark Viola Apr 23 at 3:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.