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Study the convergence of the following integral for $\alpha\in\mathbb{R}$ and evaluate it for $\alpha = \frac{1}{3}$ if it converges for that value.

$${\Large\int} _{0}^{1}\frac{dx}{(e^x-1)^{\alpha}} x$$

The integral converges for $\alpha \lt 1$, if I am not wrong, so I can evaluate: $${\Large\int} _{0}^{1}\frac{dx}{(e^x-1)^{\frac{1}{3}}}$$

whose primitive, using some razionalizations and change of variables, is: $\frac{1}{2}\log|4(e^x-1)^\frac{2}{3}-4(e^x-1)^\frac{1}{3}+4|+\sqrt{3}\arctan\left(\frac{1}{\sqrt{3}}\left(2(e^x-1)^\frac{1}{3}-1\right)\right)-\log|(e^x-1)^\frac{1}{3}+1|+C$

Anyone knows if there is a less intricate way in order of obtain it than almost 3 sheet full of calculations? Thank you

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    $\begingroup$ There seems to be an excess $x$ in the first displayed equation? $\endgroup$ – joriki Aug 10 '18 at 19:20
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    $\begingroup$ How could you prove that it converges for $\alpha <1 $? $\endgroup$ – Obvious Aug 10 '18 at 19:40
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    $\begingroup$ Will the substitution $(e^x -1)^{1/3} = u$ make it any simpler? $\endgroup$ – MathNovice Aug 10 '18 at 19:47
  • $\begingroup$ Using \Large is rarely, and certainly not here, a good idea. $\endgroup$ – Did Aug 10 '18 at 20:29
  • $\begingroup$ @Did I have used \ left( , \ right) $\endgroup$ – F.inc Aug 10 '18 at 20:33
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$$\int_{0}^{1}\frac{dx}{\sqrt[3]{e^x-1}}\stackrel{x\mapsto \log t}{=}\int_{1}^{e}\frac{dt}{t\sqrt[3]{t-1}}\stackrel{t\mapsto s+1}{=}\int_{0}^{e-1}\frac{ds}{(s-1)\sqrt[3]{s}}\stackrel{s\mapsto u^3}{=}\int_{0}^{\sqrt[3]{e-1}}\frac{3u}{u^3-1}\,du$$ where $$ \frac{3u}{u^3-1} \stackrel{\text{Residues}}{=} \frac{1}{u-1}+\frac{1-u}{u^2+u+1} $$ and $$ \frac{1-u}{u^2+u+1} = \frac{4}{(2u+1)^2+3}-\frac{4u}{(2u+1)^2+3} $$ lead to the answer in less than a page. This is a trick you are probably going to apply many times: turn the integrand function into a rational function, then perform a partial fraction decomposition. Objects like $\frac{1}{(u+\alpha)^n}$ and $\frac{1}{(u^2+\alpha^2)^n}$ always have simple primitives, that's the core of symbolic integration algorithms.

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using $u={{\left( {{e}^{x}}-1 \right)}^{1/3}}$ leads to $\int{\frac{3u}{{{u}^{3}}+1}du}$ $$ \begin{align} & =\int{-\frac{1}{u+1}+\frac{u+1}{{{u}^{2}}-u+1}du} \\ & =\int{-\frac{1}{u+1}du}+\frac{1}{2}\int{\frac{2u-1}{{{u}^{2}}-u+1}du}+\frac{3}{2}\int{\frac{1}{{{u}^{2}}-u+1}du} \\ & =\ln \left( u+1 \right)+\frac{1}{2}\ln \left( {{u}^{2}}-u+1 \right)+\frac{3}{2}\int{\frac{1}{{{u}^{2}}-u+1}du} \\ \end{align} $$ For the last integral use $v=2u-1$ and notice that $4\left( {{u}^{2}}-u+1 \right)={{v}^{2}}+3$ $$ \begin{align} & =\ln \left( u+1 \right)+\frac{1}{2}\ln \left( {{u}^{2}}-u+1 \right)+3\int{\frac{1}{{{v}^{2}}+3}dv} \\ & =\ln \left( u+1 \right)+\frac{1}{2}\ln \left( {{u}^{2}}-u+1 \right)+3\frac{1}{\sqrt{3}}ta{{n}^{-1}}\left( \frac{2u-1}{\sqrt{3}} \right) \\ \end{align} $$

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Take $$t = (e^x - 1)^{1/3}$$ or $$x = \ln( t^3 + 1 )$$ Then $$dt = \frac{1}{3}e^x(e^x-1)^{-2/3} \ dx$$ Hence $${\Large\int} _{0}^{1}\frac{dx}{(e^x-1)^{\frac{1}{3}}} = 3{\Large\int} _{0}^{(e-1)^{1/3}}\frac{1 }{t e^x(e^x-1)^{-2/3}}\ dt = 3{\Large\int} _{0}^{(e-1)^{1/3}}\frac{t}{(t^3 + 1)}\ dt $$ The last integral could be written as $$ {\displaystyle\int} _{0}^{(e-1)^{1/3}}\dfrac{t}{\left(t+1\right)\left(t^2-t+1\right)}\,\mathrm{d}t$$ Perform Partial Fraction Decomposition $${\displaystyle\int} _{0}^{(e-1)^{1/3}}\left(\dfrac{t+1}{3\left(t^2-t+1\right)}-\dfrac{1}{3\left(t+1\right)}\right)\mathrm{d}t$$ Take the first integral above, i.e. ${\displaystyle\int}\dfrac{t+1}{t^2-t+1}\,\mathrm{d}t$ could be written as $$\class{steps-node}{\cssId{steps-node-7}{\dfrac{1}{2}}}{\displaystyle\int}\dfrac{2t-1}{t^2-t+1}\,\mathrm{d}t+\class{steps-node}{\cssId{steps-node-8}{\dfrac{3}{2}}}{\displaystyle\int}\dfrac{1}{t^2-t+1}\,\mathrm{d}t$$ which is equal to (with some straight forward math) $$\dfrac{\ln\left(t^2-t+1\right)}{2}+\sqrt{3}\arctan\left(\dfrac{2t-1}{\sqrt{3}}\right)$$ Now, we need $\dfrac{\ln\left(t^2-t+1\right)}{2}+\sqrt{3}\arctan\left(\dfrac{2t-1}{\sqrt{3}}\right)$ which is easily shown to be $$\dfrac{\ln\left(t^2-t+1\right)}{2}+\sqrt{3}\arctan\left(\dfrac{2t-1}{\sqrt{3}}\right)==\ln\left(t+1\right)$$ Finalizing we get $$3\left(-\dfrac{\ln\left(\left|t+1\right|\right)}{3}+\dfrac{\ln\left(t^2-t+1\right)}{6}+\dfrac{\arctan\left(\frac{2t-1}{\sqrt{3}}\right)}{\sqrt{3}}\right)\Big\vert_0^{(e-1)^{1/3}}$$ which is $$3\left(\dfrac{\ln\left(\left(\mathrm{e}-1\right)^\frac{2}{3}-\sqrt[3]{\mathrm{e}-1}+1\right)}{6}+\dfrac{\arctan\left(\frac{2\cdot\sqrt{3}\sqrt[3]{\mathrm{e}-1}-\sqrt{3}}{3}\right)}{\sqrt{3}}-\dfrac{\ln\left(\sqrt[3]{\mathrm{e}-1}+1\right)}{3}+\dfrac{{\pi}}{2{\cdot}3^\frac{3}{2}}\right)$$

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