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Prove that: $$\int_0^\infty \frac{\ln x}{x^a+1}\;\text{d}x=-\left( \frac{\pi }{a} \right)\cot \left( \frac{\pi }{a} \right)\csc \left( \frac{\pi }{a} \right),\ \ a>1$$ For this one I consider to have $\displaystyle\int_0^\infty \ln x\int_0^\infty \text{e}^{-y(1+x^a)} \, \text{d}y \, \text{d}x$.

And use $\displaystyle{\Gamma}'(s)=\int_0^\infty \ln(x) x^{s-1} \text{e}^{-x} \, \text{d}x$, but then I just stuck...

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For what values of $a$ does this integral converge? – Ron Gordon Jan 30 '13 at 13:24
@rlgordonma Sry, for $a>1$ – gauss115 Jan 30 '13 at 13:26
up vote 8 down vote accepted

Consider a two-parameter integral: $$ f(s,a) = \int_0^\infty \frac{x^s}{x^a+1}\mathrm{d}x \stackrel{y=x^a}{=} \frac{1}{a} \int_0^\infty \frac{y^{(s+1)/a-1}}{1+y} \mathrm{d}y $$ The integral is convergent for $1>\frac{s+1}{a}>0$. You can now use your trick: $$\begin{eqnarray} f(s,a) &=& \frac{1}{a} \int_0^\infty \int_0^\infty y^{(s+1)/a-1} \exp(-t(1+y)) \mathrm{d}y \mathrm{d}t \\ &=& \frac{1}{a} \Gamma\left(\frac{s+1}{a}\right)\int_0^\infty t^{-(s+1)/a} \exp(-t) \mathrm{d}t \\ &=& \frac{1}{a} \Gamma\left(\frac{s+1}{a}\right) \Gamma\left(1-\frac{s+1}{a}\right) = \frac{\pi}{a} \frac{1}{\sin\left(\pi \frac{s+1}{a} \right)} \end{eqnarray} $$ The integral you seek to evaluate is recovered as $$ \left.\frac{\mathrm{d}}{\mathrm{d}s}f(s,a)\right|_{s=0} = -\frac{\pi^2}{a^2} \frac{1}{\sin\left(\frac{\pi}{a}\right)} \frac{1}{\tan\left(\frac{\pi}{a}\right)} $$ So it appears you are missing a power in the expected answer.

Numerical confirmation with Mathematica:

In[45]:= N[With[{a = 7/5}, -((Pi^2 Cot[Pi/a] Csc[Pi/a])/a^2)], 20]

Out[45]= 5.1362566843868779343

In[46]:= N[
 NIntegrate[Log[x]/(1 + x^(7/5)), {x, 0, Infinity}, 
  WorkingPrecision -> 25], 20]

Out[46]= 5.1362566843868779343
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Very similar to Sasha's elegant answer, just without the double integral!


Let $t=\dfrac{1}{x^a+1}:$


Using the Beta-Gamma relation:


By the lovely reflection formula:





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