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[QUOTE="julian, post: 6820738, member: 142346"] Seeing as I had already given an answer in post #3, I gave a suggestion as how to evaluate the integral via another method in post #4 (the third suggestion). Seeing as the two weeks are nearly up, I'll post how to do it this other way: [SPOILER] First use the substitution ##u = e^x## and obtain $$ I = \int_{-\infty}^\infty \frac{e^{a x}}{1 + e^x} dx = \int_0^\infty \frac{u^{a-1}}{1+u} du $$ This ##u## integral can be evaluated via a contour integral. First note that the function $$ f(z) = \frac{z^{a-1}}{1+z} $$ has a branch point at ##z=0##. We take the branch line to be the positive real axis. We employ the contour integral shown in the figure. We integrate along the positive real axis just above the branch cut, then around a large circle, then along the positive real axis in the negative direction just below the cut, but on the same branch, and then around a small circle. [ATTACH type="full" width="351px" alt="contour.jpg"]317059[/ATTACH] Just below the cut, but on the same branch, we have $$ \frac{(e^{i 2 \pi} u)^{a-1} }{1+e^{i 2 \pi} u} = e^{i 2 \pi (a-1)} \frac{u^{a-1}}{1+u} = e^{i 2 \pi a} \frac{u^{a-1}}{1+u} $$ For the integration around a small circle around the origin put ##z = \epsilon e^{i \theta}##, then \begin{align*} \left| \int_0^{2 \pi} \frac{\epsilon^a e^{i \theta a} i}{1 + \epsilon e^{i \theta}} d \theta \right| & \leq \epsilon^a \int_0^{2 \pi} \frac{d \theta}{\sqrt{\epsilon^2 + 2 \epsilon \cos \theta + 1}} \nonumber \\ & \leq \epsilon^a \int_0^{2 \pi} \frac{d \theta}{\sqrt{\epsilon^2 - 2 \epsilon + 1}} \nonumber \\ & = 2 \pi \epsilon^a \frac{1}{1 - \epsilon} \rightarrow 0 \quad \text{as} \; \epsilon \rightarrow 0 \end{align*} For integration around circle of large radius put ##z = R e^{i \theta}##, then \begin{align*} \left| \int_0^{2 \pi} \frac{R^a e^{i \theta a} i}{1 + R e^{i \theta}} d \theta \right| & \leq \int_0^{2 \pi} \frac{R^{a-1} d \theta}{\sqrt{R^{-2} + 2 R^{-1} \cos \theta + 1}} \nonumber \\ & \leq \int_0^{2 \pi} \frac{R^{a-1} d \theta}{\sqrt{R^{-2} - 2 R^{-1} + 1}} \nonumber \\ & = 2 \pi \frac{R^{a-1}}{1 - R^{-1}} \rightarrow 0 \quad \text{as} \; R \rightarrow \infty \end{align*} Therefore, $$ \oint \frac{z^{a-1}}{1+z} = \int_0^\infty \frac{u^{a-1}}{1+u} du - e^{i 2 \pi a} \int_0^\infty \frac{u^{a-1}}{1+u} du $$ The function ##f(z)## has a simple pole at ##z=-1##, so that $$ \oint \frac{z^{a-1}}{1+z} =2 \pi i (e^{i \pi})^{(a-1)} = - 2 \pi i e^{i \pi a} . $$ Combining the above gives $$ (1 - e^{i 2 \pi a}) \int_0^\infty \frac{u^{a-1}}{1+u} du = - 2 \pi i e^{i \pi a} $$ or $$ \int_0^\infty \frac{u^{a-1}}{1+u} du = \frac{\pi}{\sin a \pi} . $$ [/SPOILER] [/QUOTE]
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