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show that $$I=\int_{0}^{1}\dfrac{(x-1)dx}{(x+1)\ln{x}}=\ln{\left(\dfrac{\pi}{2}\right)}$$

I say this integral background,today I have use this wolf play some function,and Suddenly I found this interesting problem

I try solve this by hand,

$$\dfrac{x-1}{(x+1)\ln{x}}=(1-\dfrac{2}{x+1})\dfrac{1}{\ln{x}}$$ then It's ugly then $$I=\int_{0}^{1}\dfrac{1}{\ln{x}}dx-\int_{0}^{1}\dfrac{2}{\ln{x}}d\ln{(1+x)}$$

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math110
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  • Perhaps "Differentiation Under the Integral Sign" would help here? My idea is to let $I(a)=\int_0^1\dfrac{x^a-1}{(x+1)\ln x}\operatorname d!x$ and then try to compute $I'(a)$. – Akiva Weinberger Dec 07 '14 at 15:18
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    See this answer: http://math.stackexchange.com/questions/280105/evaluating-int-01-frac1-x1x-frac-mathrm-dx-ln-x/280134#280134 – Ron Gordon Dec 07 '14 at 15:30
  • Oh,Thank you very much! @RonGordon – math110 Dec 07 '14 at 15:37

1 Answers1

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We have: $$ I = \int_{0}^{+\infty}\frac{1-e^{-x}}{1+e^{x}}\frac{dx}{x} = \int_{0}^{+\infty}\int_{0}^{+\infty}\frac{1-e^{-x}}{1+e^{x}}e^{-tx}\,dt\,dx$$ So: $$ I = \frac{1}{2}\int_{0}^{+\infty}\left(-H_{(t-1)/2}+2H_{t/2}-H_{(t+1)/2}\right)\,dt = \frac{1}{2}\int_{0}^{1}\left(H_{t/2}-H_{(t-1)/2}\right)\,dt.$$ Once one proves that: $$\int_{0}^{1/2}H_t\,dt = \frac{1}{2}\left(\gamma+\log\frac{\pi}{4}\right),$$ $$\int_{-1/2}^{0}H_t\,dt = \frac{1}{2}\left(\gamma-\log\pi\right),$$ the result follows. The previous identities are equivalent to: $$2\int_{0}^{1/2}\psi(t+1)\,dt = \log\frac{\pi}{4},$$ $$2\int_{-1/2}^{0}\psi(t+1)\,dt = -\log\pi,$$ but since $\psi=\frac{d}{dz}\log\Gamma(z)$, we have: $$\int_{0}^{1/2}\psi(t+1)\,dt = \log\Gamma\left(\frac{3}{2}\right)-\log\Gamma\left(1\right),$$ $$\int_{-1/2}^{0}\psi(t+1)\,dt = \log\Gamma(1)-\log\Gamma\left(\frac{1}{2}\right),$$ and we are done.

Jack D'Aurizio
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