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$$ \int_{0}^{\infty } \frac{ln x}{x^{3/4}(1+x)}dx$$

Integrating the real axis from 0 to infinity, so go around a big circle and close from - infinity to 0, around 0 close the path with a little circle.

$x = r$ along the real axis and Re>0. $I = \int_{0}^{\infty } \frac{ln x}{x^{3/4}(1+x)}dx$

Along the big circle, the integral vanishes.

along the real axis and Re>0, $ \int_{\infty}^{0} \frac{e^{i \pi /2 }ln r}{r^{3/4}(1+r)}dr = - \int_{0}^{\infty} \frac{e^{i \pi /2 }ln r}{r^{3/4}(1+r)}dr = -iI$

Along the little circle it will diverge ...

Along a circle around minus one, it diverges too. So what should i do? I can't see any other point .

Lac
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2 Answers2

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If you're willing to use Fourier series instead of complex analysis, we can derive a more powerful result.

In this question, we use Fourier series to show that for $0 < \alpha < 1$ $$I(\alpha)=\int_0^{+\infty} \frac {1}{x^{\alpha}(1+x)}dx = \frac \pi {\sin(\pi \alpha)}$$ So taking the derivative w.r.t. $\alpha$ yields, for $0<\alpha < 1$, $$\boxed{\int_0^{+\infty} \frac {\ln(x)}{x^{\alpha}(1+x)}dx = \frac {\pi^2\cos(\pi \alpha)} {\sin^2(\pi \alpha)}}$$ Evaluating at $\alpha = \frac 3 4$ gives $$\boxed{\int_0^{+\infty} \frac {\ln(x)}{x^{\frac 3 4}(1+x)}dx = -\pi^2 \sqrt{2}}$$

Stefan Lafon
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There is another possible approach based on the fact than we have the antidericative $$\int \frac{ \log (x)}{x^{a}(x+1)}\,dx=$$ $$-\frac{x^{1-a}}{(1-a)^2}\Bigg[\, _3F_2(1,1-a,1-a;2-a,2-a;-x)+(a-1) \log (x) \, _2F_1(1,1-a;2-a;-x) \Bigg]$$ which could also write $$-x^{-a} \left(x \Phi (-x,2,1-a)+(-x)^a \log (x) B_{-x}(1-a,0)\right)$$

Using the integration bounds $$\int_0^\infty \frac{ \log (x)}{x^{a}(x+1)}\,dx=\pi ^2 \cot (\pi a) \csc (\pi a)\qquad \text{if} \qquad 0 <a <1$$

Claude Leibovici
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