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$$\begin{align} & \int_{0}^{\frac{1}{2}}{\frac{\ln x{{\ln }^{2}}\left( 1-x \right)}{x}}\text{d}x \\ & \int_{0}^{+\infty }{\tan \frac{x}{\sqrt{{{\pi }^{2}}+{{x}^{3}}}}\frac{\ln \left( 1+\sqrt{x} \right)}{x}}\text{d}x \\ \end{align}$$

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8  
Yes, those are some integrals with logarithms appearing in them. –  Zev Chonoles Feb 4 '13 at 4:33
3  
A comment. ${}{}{}$ –  mixedmath Feb 4 '13 at 4:34
    
The first one's $\frac{dx}{x}$ suggests trying substituting $u = \ln x$ and see what happens. Perhaps the $\ln (1 - x)$ succumbs to symmetry around $x = 1 / 2$ later... In the second it is also tempting, but I don't think that will work. Are complex analysis tools allowed? –  vonbrand Feb 4 '13 at 5:34
    
@vonbrand : Yes, any approaches would be fine –  Ryan Feb 4 '13 at 5:49

1 Answer 1

up vote 1 down vote accepted

A similar technique was used to solve a previous problem. We give a solution for the first integral. Integration by parts with $ u = \ln(1-x)^2 $ gives

$$ \int_{0}^{\frac{1}{2}}{\frac{\ln x{{\ln }^{2}}\left( 1-x \right)}{x}}\text{d}x = \frac{\ln^4(2)}{2}+\int_{0}^{\frac{1}{2}}{\frac{\ln^2 x\,{{\ln }}\left( 1-x \right)}{1-x}}\text{d}x $$

$$ =\frac{\ln^4(2)}{2} -\sum_{n=0}^{\infty}H_{n}\int_{0}^{\frac{1}{2}}x^n\ln^2(x) dx$$

$$ = \frac{\ln^4(2)}{2}-\sum_{n=1}^{\infty}\frac{H_{n-1}}{2^n}\left( \frac{\ln^2(2)}{n}+\frac{2\ln(2)}{n^2}+ \frac{2}{n^3} \right)\sim -0.2128720338,$$

where $H_n$ are the harmonic numbers. Note that, we used the power series expansion of the function $ \frac{\ln(1-x)}{1-x}$

$$ -\sum_{n=0}^{\infty}H_nx^n .$$

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