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If one performs in the improper integral
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(1) |
the change of variable $$x \;=\; -\ln{t}, \quad dx = -\frac{dt}{t},$$ 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
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(2) |
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 method of differentiation under the integral sign or
taking Laplace transform with respect to $k$ yields a more complicated integral.
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