# example of changing variable

## Primary tabs

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\begin{document}
If one performs in the improper integral
\begin{align}
I \;:=\; \int_{-\infty}^\infty\frac{e^{kx}}{1\!+\!e^x}\,dx \qquad (0 < k < 1)
\end{align}
$$x \;=\; -\ln{t}, \quad dx = -\frac{dt}{t},$$
the new lower limit becomes $\infty$ and the new upper limit 0; hence one obtains
$$I \;=\; -\int_\infty^0\frac{e^{-k\ln{t}}dt}{(1\!+\!e^{-\ln{t}})t} \;=\; \int_0^\infty\frac{t^{-k}}{t\!+\!1}\,dt.$$
Thus one has recurred $I$ to the integral
\begin{align}
\int_0^\infty\frac{x^{-k}}{x\!+\!1}\,dx,
\end{align}
the value of which has been determined in the entry using residue theorem near branch point.\, Accordingly, we may write the result
$$\int_{-\infty}^\infty\frac{e^{kx}}{1\!+\!e^x}\,dx \;=\; \frac{\pi}{\sin{\pi k}}.$$\\

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
$$-a \,\leqq\, \mbox{Re}\,z \,\leqq\, a, \quad 0 \,\leqq\, \mbox{Im}\,z \,\leqq\, 2\pi$$
and then to let\, $a \to \infty$ (one cannot use the same half-disk as in determining the integral (2)).\, As for using the \PMlinkname{method}{MethodsOfEvaluatingImproperIntegrals} of differentiation under the integral sign or taking Laplace transform with respect to $k$ yields a more complicated integral.

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