Evaluating $\int_{0}^{\infty}\frac{\ln\left(x\right)\left(e^{-x}-\cos x\right)}{x}dx$

Question

From here after the partial fractions in the question :

$$I = \int_{0}^{\infty}\frac{\ln\left(x\right)\left(e^{-x}-\cos x\right)}{x}dx$$

I tried this via contour integration but ran into some problems with the choice of contours and the convergence of the circular arcs. I'm not sure if I need a branch cut yet, but that doesn't affect this integral. Going anticlockwise and having $z=Re^{i\theta}$ :

$$\begin{align}\int_{\Gamma}&=\int_{0}^{\pi}{\frac{\ln(Re^{i\theta})(e^{-Re^{i\theta}}-e^{iRe^{i\theta}})}{Re^{i\theta}}iRe^{i\theta}\, d\theta}\\&=i\int_{0}^{\pi}(\ln(R)+i\theta)(e^{-Re^{i\theta}}-e^{iRe^{i\theta}})\,d\theta\end{align}$$

But I don't think this converges given $R\to\infty$. Should a different contour be used or can this be continued?

Answer 1

By Frullani's theorem $$ \ln(b)=\int_{0}^{+\infty}\frac{e^{-x}-e^{-bx}}{x}\,dx \tag{1}$$ holds for any $b\in\mathbb{C}$ in the right half-plane ($\operatorname{Re}(b)>0$). Under the same assumptions, the slight generalization $$\Gamma(k)\left(1-\frac{1}{b^k}\right) = \int_{0}^{+\infty} x^k\frac{e^{-x}-e^{-bx}}{x}\,dx \tag{2} $$ holds for any $k>0$. By differentiating both sides of $(2)$ with respect to $k$, then considering the limit as $k\to 0^+$, $$ -\frac{1}{2}\ln(b)(2\gamma+\ln(b)) = \int_{0}^{+\infty}\ln(x)\frac{e^{-x}-e^{-bx}}{x}\,dx. \tag{3}$$ By considering the limit of both sides when $b$ approaches $i$ from the right half-plane we get $$ \int_{0}^{+\infty}\ln(x)\frac{e^{-x}-e^{-ix}}{x}\,dx = \frac{\pi^2}{8}-\frac{\pi i}{2}\gamma \tag{4}$$ and finally, by considering the real parts of both sides, $$ \int_{0}^{+\infty}\ln(x)\frac{e^{-x}-\cos x}{x}\,dx = \frac{\pi^2}{8}.\tag{5} $$ As usual, the properties of the (inverse) Laplace transform allow us to avoid the hunt for a suitable contour.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\iverson}[1]{\left[\left[\,{#1}\,\right]\right]} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#44f}{\LARGE\tt I} & \equiv\color{#44f}{\int_{0}^{\infty}{\ln\left(x\right) \bracks{\expo{-x} - \cos\pars{x}} \over x}\dd x} \\[5mm] & = \lim_{\epsilon\ \to\ 0^{+}}\,\, \partiald{}{\epsilon}\int_{0}^{\infty} x^{\epsilon - 1}\bracks{\expo{-x} - \cos\pars{x}}\dd x \\[5mm] & = \lim_{\epsilon\ \to\ 0^{+}}\,\, \partiald{}{\epsilon}\bracks{\int_{0}^{\infty} x^{\epsilon - 1}\expo{-x}\dd x - {1 \over 2}\int_{0}^{\infty} x^{\epsilon/2 - 1}\cos\pars{\root{x}}\dd x} \end{align} Note that $ \ds{\color{red}{\cos\pars{\root{x}}} = \sum_{n = 0}^{\infty}\color{red}{\Gamma\pars{1 + n} \over \Gamma\pars{1 + 2n}}\,{\pars{-x}^{n} \over n!}} $. With the $\ds{Ramanujan's\ Master\ Theorem}$: \begin{align} \color{#44f}{\LARGE\tt I} & \equiv\color{#44f}{\int_{0}^{\infty}{\ln\left(x\right) \bracks{\expo{-x} - \cos\pars{x}} \over x}\dd x} \\[5mm] & = \lim_{\epsilon\ \to\ 0^{+}}\,\, \partiald{}{\epsilon}\bracks{\Gamma\pars{\epsilon}\color{red}{1} - {1 \over 2}\Gamma\pars{\epsilon \over 2}\color{red}{\Gamma\pars{1 - \epsilon/2} \over \Gamma\pars{1 - \epsilon}}} \\[5mm] & = \lim_{\epsilon\ \to\ 0^{+}}\,\, \partiald{}{\epsilon}\bracks{\Gamma\pars{\epsilon} - {1 \over 2}{\pi \over \sin\pars{\pi\epsilon/2}}{\Gamma\pars{\epsilon} \over \pi/\sin\pars{\pi\epsilon}}} \\[5mm] & = \lim_{\epsilon\ \to\ 0^{+}}\,\, \partiald{}{\epsilon}\braces{\Gamma\pars{\epsilon}\bracks{1 - \cos\pars{\pi\epsilon \over 2}}} \\[5mm] & = 2\lim_{\epsilon\ \to\ 0^{+}}\,\, \partiald{}{\epsilon}\bracks{\Gamma\pars{\epsilon} \sin^{2}\pars{\pi\epsilon \over 4}} \\[5mm] & = 2\lim_{\epsilon\ \to\ 0^{+}}\,\, \partiald{}{\epsilon}\pars{{1 \over \epsilon} {\pi^{2}\epsilon^{2} \over 16}} = \bbx{\color{#44f}{\pi^{2} \over 8}} \approx 1.2337 \\ & \end{align}

Answer 3

Consider \begin{align} I(a)&=\int_{0}^{\infty}\frac{\ln x}x\left[e^{-x}(1-\cos ax )+(e^{-ax}-1)\cos x\right] dx\\ I’(a)&= \int_0^\infty \ln x\left( e^{-x}\sin ax -e^{-ax}\cos x\right)dx =\frac{\pi}{2(1+a^2)} \end{align} Then $$\int_{0}^{\infty}\frac{\ln x\left(e^{-x}-\cos x\right)}{x}dx=I(1)=\int_0^1 I’(a) da=\frac{\pi^2}8 $$

Answer 4

If you use the complex integration, it is convenient to consider the integrand $\displaystyle f(z)=\ln z\,\frac{e^{-z}-e^{iz}}z$ and closed contour $$r\to R\to iR \,(\text{along the arch of a big radius R, counter-clockwise})$$ $$\to ir\to r\,(\text{along the arch of a small radius r, clockwise})$$ and the cut along the positive part of the axis X (to make $\ln z $ a single-valued function). We stay on the upper bank of the cut.

Denoting the integral along the arches as $I_{C_R}$ and $I_{C_r}$, $$\oint f(z)dz=\int_r^R\ln x\,\frac{e^{-x}-e^{ix}}xdx+I_{C_R}+\int_{iR}^{ir}\ln x\,\frac{e^{-x}-e^{ix}}xdx+I_{C_r}=0$$ because we do not have poles inside the contour. Using $\,\displaystyle x=e^\frac{\pi i}2t\,$ for the second integral, $$\int_r^R\ln x\,\frac{e^{-x}-e^{ix}}xdx+\int_r^R\ln x\,\frac{e^{-x}-e^{-ix}}xdx=2\int_r^R\ln x\,\frac{e^{-x}-\cos x}xdx$$ $$=\frac{\pi i}2\int_r^R\frac{e^{-ix}-e^{-x}}xdx-I_{C_R}-I_{C_r}$$ Integral along small and big arches tend to zero.

Indeed, $$I_{C_r}=\int_0^{\pi/2}\ln(re^{i\phi})\frac{e^{-re^{i\phi}}-e^{ire^{i\phi}}}{re^{i\phi}}ire^{i\phi}d\phi=O(r\ln r)\to0 \,\text{at}\,r\to0$$ $$I_{C_R}=\int_0^{\pi/2}\ln(Re^{i\phi})\frac{e^{-Re^{i\phi}}-e^{iRe^{i\phi}}}{Re^{i\phi}}iRe^{i\phi}d\phi=i\int_0^{\pi/2}\ln(Re^{i\phi})\left(e^{-Re^{i\phi}}-e^{iRe^{i\phi}}\right)d\phi$$ where every integral can be estimated by means of Jordan's lemma. For example, $$\Big|I_1\Big|=\Big|i\int_0^{\pi/2}\ln(Re^{i\phi})e^{-Re^{i\phi}}d\phi\Big|<\ln R\int_0^{\pi/2}e^{-R\cos\phi}d\phi=\ln R\int_0^{\pi/2}e^{-R\sin\phi}d\phi$$ Using $\sin\phi\geqslant \frac2\pi\phi$ for $\phi\in[0;\frac\pi2]$ $$\Big|I_1\Big|<\ln R\int_0^{\pi/2}e^{-\frac2\pi R\phi}d\phi=\frac\pi2\frac{\ln R}R\left(1-e^{-R}\right)\to 0\,\text{at}\,R\to\infty$$ Therefore, leading $r\to 0;\,R\to\infty$ $$2\int_0^\infty\ln x\,\frac{e^{-x}-\cos x}xdx=\frac{\pi i}2\int_0^\infty\frac{e^{-ix}-e^{-x}}xdx$$ $$=\frac\pi2\int_0^\infty\frac{\sin x}xdx+\frac{\pi i}2\int_0^\infty\frac{\cos x-e^{-x}}xdx$$ $$=\frac{\pi^2}4+\frac{\pi i}2\int_0^\infty\frac{\cos x-e^{-x}}xdx$$ The second integral is zero - the approach is exactly the same as for the initial integral (by means of a quarter-circle in the complex plane). But, in fact, we do not need to evaluate it: the initial integral is real, and the last integral is imaginary, so it must equals zero. Hence, $$\int_0^\infty\ln x\,\frac{e^{-x}-\cos x}xdx=\frac{\pi^2}8$$