example of changing variable

If one performs in the improper integral

(1)

the change of variable (http://planetmath.org/ChangeOfVariableInDefiniteIntegral)

$$x=-\ln t,dx=-\frac{dt}{t},$$

the new lower limit becomes $\mathrm{\infty }$ and the new upper limit 0; hence one obtains

$$I=-{\int }_{\mathrm{\infty }}^0\frac{e^{-k\ln t}dt}{(1+e^{-\ln t})t}={\int }_0^{\mathrm{\infty }}\frac{t^{-k}}{t+1}𝑑t.$$

Thus one has recurred $I$ to the integral

${\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle \frac{x^{-k}}{x+1}}𝑑x,$(2)

the value of which has been determined in the entry using residue theorem near branch point.  Accordingly, we may write the result

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{kx}}{1+e^x}𝑑x=\frac{\pi }{\sin \pi k}.$$

Calculating the integral (1) directly is quite laborious:  one has to use Cauchy residue theorem to the integral

$${\oint }_c\frac{e^{kz}}{1+e^z}𝑑z$$

about the perimetre $c$ of the rectangle

$$-a\leqq \text{Re}z\leqq a,0\leqq \text{Im}z\leqq \mathrm{\hspace{0.17em}2}\pi$$

and then to let  $a\to \mathrm{\infty }$ (one cannot use the same half-disk as in determining the integral (2)).  As for using the method (http://planetmath.org/MethodsOfEvaluatingImproperIntegrals) of differentiation under the integral sign or taking Laplace transform with respect to $k$ yields a more complicated integral.