For all $-1<k<1$, I'm trying to solve the following integral: $$\int_{-\infty}^\infty z\frac{e^{z(k+1)}}{(1+e^z)^2}dz.$$ My firt attempt was to use change of variables: $t=e^z\implies z=\ln(t)\implies dz/dt=1/t$. Hence, \begin{equation} \int_{-\infty}^\infty z\frac{e^{z(k+1)}}{(1+e^z)^2}dz=\int_{0}^\infty \ln(t)\frac{t^{k}}{(1+t)^2}dt. \quad (1) \end{equation} But the expression on the RHS of $(1)$ is not easy to solve. WolframAlpha asserts that \begin{equation} \int_{0}^\infty \ln(t)\frac{t^{k}}{(1+t)^2}dt=\pi(1-\pi k \cot(\pi k))\csc(\pi k), \quad (2) \end{equation} which I'm unable to show.

Can someone help me with the steps to show that $(2)$ is valid?

  • 2
    $\begingroup$ Just a hunch, maybe some residues may be useful, as I see cotangents and cosecants of certain angles. $\endgroup$ Commented yesterday

2 Answers 2


There's something really clever at play here using the beta function. Using the second formula as seen in the Wikipedia article (which uses the substitution $t \mapsto \frac{t}{1+t}$; see, e.g. here, we can write

$$B(p, q) = \int_0 ^1 t^{p-1} (1-t)^{q-1} \ dt = \int_0 ^\infty \frac{t^{p+q-1}}{(1+t)^{p+q}} \ dt. $$

Set $p = k+1$ and $q = 1-k.$ We have the related integral

$$B(k+1, 1-k) = \int_0 ^\infty \frac{t^k}{(1+t)^2} \ dt$$

which is very close to the desired integral. Now, as $$\frac{\partial}{\partial k} t^k = t^k \ln t$$ we'll follow the trick from Svyatoslav so we obtain

$$ \frac{\partial B(k+1, 1-k)}{\partial k} = \int_0 ^\infty \ln t \frac{t^k}{(1+t)^2} \ dt.$$

Now by using properties of the beta and gamma functions,

\begin{align*} B(k+1, 1-k) &= \frac{\Gamma(k+1) \Gamma(1-k)}{\Gamma(2)} \\ &= k\Gamma(k) \Gamma(1-k) \\ &= \frac{k\pi}{\sin k \pi}. \end{align*}

Taking the derivative now gives

$$\frac{d}{dk} \left( \frac{k\pi}{\sin k \pi} \right) = \pi \csc k\pi \left( 1 - k\pi \cot k\pi \right)$$

as desired.

  • $\begingroup$ What did you use to ensure the interchange between improper integral and differentiation? $\endgroup$ Commented 20 hours ago
  • $\begingroup$ Note that $t^k$ is continuous for $k \in (-1, 1)$ as well as its derivative. $\endgroup$ Commented 20 hours ago

Let’s start with $$ \int_{-\infty}^{\infty} \frac{z e^{z(k+1)}}{\left(1+e^z\right)^2} d y \stackrel{t=e^z}{=} \int_0^{\infty} \frac{t^k \ln t}{(1+t)^2} d t=I’(k) $$ where the parametrised integral $$ I(a)=\int_0^{\infty} \frac{t^a}{(1+t)^2} d t. $$ Let $y=\frac{1}{1+t} $, then $$ \begin{aligned} I(a) & =\int_0^1\left(\frac{1}{y}-1\right)^a d y \\ & =\int_0^1 y^{-a}(1-y)^a d y \\ & =B(1-a, 1+a) \\ & =\frac{\Gamma(1-a) \Gamma(1+a)}{\Gamma(2)} \\ & =a \Gamma(1-a) \Gamma(a) \\ & =a \pi \csc (\pi a) \end{aligned} $$ Differentiating $I(a)$ w.r.t. $a$ at $a=k$ yields $$ \boxed{\int_{-\infty}^{\infty} \frac{z e^{z(k+1)}}{\left(1+e^z\right)^2} d y = I^{\prime}(k)=\csc (\pi k)(1-k \pi \cot (\pi k))} $$


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