$$\mathcal{J}:=\int_0^\infty \log(1-e^{-a x})\cos (bx)\, dx=\frac{a}{2b^2}-\frac{\pi}{2b}\coth \frac{\pi b}{a},\qquad \mathcal{Re}(a)>0, b>0. $$ I tried to write $$ \mathcal{J}=-\int_0^\infty \sum_{n=1}^\infty\frac{e^{-anx}}{n}\cos(bx)\,dx $$ but the taylors series, $\log (1-\xi)=-\sum_{n=1}^\infty \xi^n/n, \ |\xi|<1$, thus this is not so useful for doing the integral. I tried to also write $$ \mathcal{J}=\frac{1}{b}\int_0^\infty \log(1-e^{-ax})d(\sin bx)=\frac{1}{b}\left(\log(1-e^{-ax})\sin (bx)\big|^\infty_0 -a\int_0^\infty \frac{\sin (bx)}{{e^{ax}-1}}dx \right), $$ the boundary term vanishes so we have $$ \mathcal{J}=\frac{a}{b}\int_0^\infty \frac{\sin(bx)}{1-e^{ax}}dx=\frac{a}{b}\mathcal{Im}\bigg[\int_0^\infty \frac{e^{ibx}}{e^{ax}-1}dx\bigg] $$ which I am not sure how to solve. Notice there are singularities at $x=2i\pi n/a, \ n\in \mathbb{Z}$.

We need to calculate the residue for all the singularities along the imaginary axis. The residue contribution to the integral $$ 2\pi i\cdot \sum_{n= 0}^\infty \frac{ e^{-2\pi nb/a}}{e^{2i \pi n}}=2\pi i \sum_{n=0}^\infty e^{n( -2\pi b/a-2i\pi)}=\frac{2\pi i}{e^{-(2\pi b/a+2\pi i)}}$$ Taking the imaginary part gives and re-writing the integral gives a different result. Where did I go wrong? How can we calculate this? Thanks


4 Answers 4


Consider the integral \begin{align} I = \int_{0}^{\infty} \ln(1-e^{-ax}) \ \cos(bx) \ dx. \end{align} Expand the logarithm to obtain \begin{align} I &= - \sum_{n=1}^{\infty} \frac{1}{n} \ \int_{0}^{\infty} e^{-a n x} \ \cos(bx) \ dx \\ &= - \sum_{n=1}^{\infty} \frac{1}{n} \ \frac{an}{ a^{2} n^{2} + b^{2} } \\ &= - \frac{1}{a} \ \sum_{n=1}^{\infty} \frac{1}{n^{2} + (b/a)^{2}}. \end{align} Using the expansion \begin{align} \coth(\pi x) = \frac{1}{\pi x} + \frac{2 x}{\pi} \sum_{n=1}^{\infty} \frac{1}{n^{2} + x^{2}} \end{align} then the value of the integral becomes \begin{align} \int_{0}^{\infty} \ln(1-e^{-ax}) \ \cos(bx) \ dx = \frac{a}{2 b^{2}} - \frac{\pi}{2 b} \ \coth\left(\frac{b \pi}{a}\right). \end{align}


This may not be the easiest method, but you appear to be interested in a contour way of going about it.

starting from your $$-a/b\int_{0}^{\infty}\frac{\sin(bx)}{e^{ax}-1}dx$$

Consider the function$$f(z)=\frac{e^{ibz}}{e^{az}-1}$$

Use a rectangle in the first quadrant with height $\displaystyle 2\pi i/a$ with quarter circle indents around $2\pi i/a$ and $0$.

There will be 6 portions to put together:


The integral can be set to 0 because there are no poles inside the contour.

Along bottom horizontal on x axis: $$I_{1}=\int_{\epsilon}^{R}\frac{e^{ibx}}{e^{ax}-1}dx$$

up right vertical side:

$$\left|\frac{e^{ibR}}{e^{aR}-1}\right|\to 0, \;\ as \;\ R\to \infty$$


along top horizontal: $$I_{3}=-\int_{\epsilon}^{R}\frac{e^{ib(x+2\pi i/a)}}{e^{a(x+2\pi i/a)}-1}dx=-e^{-2\pi b/a}\int_{\epsilon}^{r}\frac{e^{ibx}}{e^{ax}-1}dx$$

top quarter circle around indent at $2\pi i/a$,

where x varies from

$(\epsilon, \epsilon+\frac{2\pi i}{a})$ to $(0,\frac{2\pi i}{a}-\frac{2\pi i}{a}\epsilon)$

$$I_{4}=\frac{-\pi i}{2}Res\left(f(z), \frac{2\pi a}{b}\right)=\frac{-\pi i}{2}\cdot \frac{e^{ib(2\pi i/a)}}{ae^{a(2\pi i/a)}}=\frac{-\pi i}{2a}e^{-2\pi b/a}$$

Down left vertical side. parameterize with $\displaystyle z=iy, \;\ dz=idy$


Quarter circle indent around the origin with x varying from $\displaystyle (0,i\epsilon)$ to $\displaystyle (\epsilon, 0)$.

$$I_{6}=\frac{-\pi i}{2}Res(f,0)=\frac{-\pi i}{2}\cdot \frac{e^{ib(0)}}{ae^{a(0)}}=\frac{-\pi i}{2a}$$

Now, assemble all the portions by plugging them all into (1):, and let $\displaystyle \epsilon\to 0, \;\ R\to \infty$

$$\int_{C}\frac{e^{ibz}}{e^{az}-1}dz=\int_{0}^{\infty}\frac{e^{ibx}}{e^{ax}-1}dx+I_{2}-e^{-2\pi b/a}\int_{0}^{\infty}\frac{e^{ibx}}{e^{ax}-1}dx$$ $$-\frac{\pi i}{2a}e^{-2\pi b/a}-\frac{\pi i}{2a}-i\int_{0}^{2\pi /a}\frac{e^{-by}}{e^{ayi}-1}dy=0$$

$$\rightarrow (1-e^{-2\pi b/a})\int_{0}^{\infty}\frac{\sin(bx)}{e^{ax}-1}dx+\int_{0}^{2\pi/a}\frac{(-i)e^{-by}}{e^{ayi}-1}dy=\frac{\pi i}{2a}(1+e^{-2\pi b/a})$$

By taking imaginary parts, the last integral(the one going down the left vertical side) can be shown to be equal to

$$\int_{0}^{2\pi/a}\frac{e^{-by}}{2}dy=\frac{1-e^{-2\pi b/a}}{2b}$$

solving for the integral in question, we finally have:

$$\int_{0}^{\infty}\frac{\sin(bx)}{e^{ax}-1}dx=\frac{\frac{\pi}{2a}(1+e^{-2\pi b/a})-\frac{1-e^{-2\pi b/a}}{2b}}{1-e^{-2\pi b/a}}$$

$$=\frac{\pi}{2a}\coth(\frac{\pi b}{a})-\frac{1}{2b}$$

multiplying this by the $-a/b$ from the beginning reduces it to a form that can be written in terms of hyperbolic trig functions as the solution suggests.

and, we ultimately get:

$$\frac{a}{2b^{2}}-\frac{\pi}{2b}\coth(\frac{\pi b}{a})$$

  • $\begingroup$ I checked this as the answer because you explained where my method failed and used this contour method. +1 Thanks $\endgroup$ May 25, 2014 at 16:41

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{{\cal J}\equiv\int_{0}^{\infty}\ln\pars{1 - \expo{-ax}}\cos\pars{bx}\, \dd x={a \over 2b^{2}} - {\pi \over 2b}\,\coth\pars{\pi b \over a}:\ {\large ?}.\quad a > 0,\ b > 0}$.

\begin{align} {\cal J}&\equiv \Re\int_{0}^{\infty}\ln\pars{1 - \expo{-ax}}\expo{\ic bx}\,\dd x =\Re\int_{0}^{\infty}\sum_{n = 1}^{\infty}\pars{-\,{\expo{-nax} \over n}} \expo{\ic bx}\,\dd x \\[3mm]&=-\,\Re\sum_{n = 1}^{\infty}{1 \over n}\int_{0}^{\infty} \expo{-\pars{na - \ic b}x}\,\dd x =-\,\Re\sum_{n = 1}^{\infty}{1 \over n}\,{1 \over na - \ic b} \\[3mm]&=-\,{1 \over a}\Re\sum_{n = 0}^{\infty} {1 \over \pars{n + 1}\pars{n + 1 - \ic b/a}} =-\,{1 \over a}\, \Re\bracks{\Psi\pars{1} - \Psi\pars{1 - \ic b/a} \over 1 - \pars{1 - \ic b/a}} \end{align} where $\ds{\Psi\pars{z}}$ is the Digamma Function ${\bf\mbox{6.3.1}}$.

\begin{align} {\cal J}&=-\,{1 \over b}\,\Im\bracks{\Psi\pars{1} - \Psi\pars{1 - {b \over a}\,\ic}} ={1 \over b}\,\Im\Psi\pars{1 - {b \over a}\,\ic} \end{align}

With the identity ${\bf\mbox{6.3.13}}$: $$ {\cal J}={1 \over b}\braces{% -\,{1 \over 2\pars{-b/a}} + \half\,\pi\coth\pars{\pi\bracks{-\,{b \over a}}}} $$

$$\color{#00f}{\large% {\cal J}\equiv\int_{0}^{\infty}\ln\pars{1 - \expo{-ax}}\cos\pars{bx}\,\dd x ={a \over 2b^{2}} - {\pi \over 2b}\,\coth\pars{\pi b \over a}} $$

  • $\begingroup$ @Integrals You're welcome. Thanks. $\endgroup$ May 25, 2014 at 21:45

Let $$ I(a)=\int_{0}^{\infty}\ln(1-e^{-ax})\cos(bx)dx. $$ Then \begin{eqnarray} I'(a)=&=&\int_{0}^{\infty}\frac{xe^{-ax}}{1-e^{-ax}}\cos(bx)dx\\ &=&\int_{0}^{\infty}\sum_{n=0}^\infty xe^{-a(n+1)x}\cos(bx)dx\\ &=&\sum_{n=0}^\infty \frac{(a(n+1)-b)(a(n+1)+b)}{(a^2(n+1)^2+b^2)^2}\\ &=&\sum_{n=1}^\infty \left(\frac{1}{a^2n^2+b^2}-\frac{2b^2}{(a^2n^2+b^2)^2}\right)\\ &=&\frac{1}{2}\left(\frac{1}{b^2}-\frac{\pi^2}{a^2\sinh^2(\frac{b\pi}{a})}\right). \end{eqnarray} So $$ I(a)=\frac{a}{2b^2}-\frac{\pi\cot(\frac{b\pi}{a})}{2b}. $$


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.