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Evaluate : $$\begin{align} & \int_{0}^{\frac{\pi }{2}}{\frac{{{\ln }^{2}}\left( 2\cos x \right)}{{{\ln }^{2}}\left( 2\cos x \right)+{{x}^{2}}}}\text{d}x \\ & \int_{0}^{1}{\frac{\arctan \left( {{x}^{3+\sqrt{8}}} \right)}{1+{{x}^{2}}}}\text{d}x \\ \end{align}$$

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For the second one, $x = \tan\theta$. – hjpotter92 Jan 29 '13 at 1:33
0.683037, 0.0763652 – Tunococ Jan 29 '13 at 1:34
For the first integral, refer to my blog posting. – Sangchul Lee Jan 29 '13 at 3:02
@sos440 : Thx sos440 :) – gauss115 Jan 29 '13 at 3:11
The first integral (subtracted from $\pi/2$) appeared before in this question. – David Moews Jan 30 '13 at 2:07
up vote 3 down vote accepted

For the second integral, consider the more general form

$$\int_0^1 dx \: \frac{\arctan{x^{\alpha}}}{1+x^2}$$

(I do not understand what is special about $3+\sqrt{8}$.)

Taylor expand the denominator and get

$$\begin{align} &=\int_0^1 dx \: \arctan{x^{\alpha}} \sum_{k=0}^{\infty} (-1)^k x^{2 k} \\ &= \sum_{k=0}^{\infty} (-1)^k \int_0^1 dx \: x^{2 k} \arctan{x^{\alpha}} \end{align}$$

Now we can simply evaluate these integrals in terms of polygamma functions:

$$\int_0^1 dx \: x^{2 k} \arctan{x^{\alpha}} = \frac{\psi\left(\frac{a+2 k+1}{4 a}\right)-\psi\left(\frac{3 a+2 k+1}{4 a}\right)+\pi }{8 k+4}$$


$$\psi(z) = \frac{d}{dz} \log{\Gamma{(z)}}$$

and we get that

$$\int_0^1 dx \: \frac{ \arctan{x^{\alpha}}}{1+x^2} = \frac{\pi^2}{16} - \frac{1}{4} \sum_{k=0}^{\infty} (-1)^k \frac{\psi\left(\frac{3 \alpha+2 k+1}{4 \alpha}\right)-\psi\left(\frac{\alpha+2 k+1}{4 \alpha}\right) }{2 k+1} $$

This is about as close as I can get. The sum agrees with the numerical integration out to 6 sig figs at about $10,000$ terms for $\alpha = 3+\sqrt{8}$.

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Thx rlgordonma, I got about the same form of answer with another log integral..which I cant simplify too :) – gauss115 Jan 29 '13 at 7:18

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