Seeking methods to solve: $\int_{0}^{1} \frac{1}{1 + \arctan(x)} \:dx$ I've been playing with the following definite integral and was wondering if anyone knew of any methods to solve?
$$I = \int_{0}^{1} \frac{1}{1 + \arctan(x)} \:dx$$
 A: The approach I took:
First let $u = \tan(x)$ to yield:
$$I = \int_{0}^{\frac{\pi}{4}} \frac{\tan^2(u) + 1}{1 + u} \:du$$
Unfortunately I had no luck with my usual tactics, so I decided to use the Taylor series of $\tan^2(u)$ at $u = 0$ which I was greatly helped with here. 
We find that:
$$ \tan^{2}(u) + 1 = \sum^{\infty}_{n=1} \frac{B_{2n} (-4)^n \left(1-4^n\right)\left(2n - 1\right)}{(2n)!} u^{2n-2} $$
Which holds for $|u| < \frac{\pi}{2}$. As the domain of the integral is within that, we can use this expansion (I believe). 
Hence, we arrive at, 
$$I = \int_{0}^{\frac{\pi}{4}} \frac{\sum_{n = 1}^{\infty} C_n u^{2n - 2}}{1 + u} \:du = \sum_{n = 1}^{\infty}C_n\int_{0}^{\frac{\pi}{4}} \frac{u^{2n - 2}}{1 + u} \:du = \sum_{n = 1}^{\infty}C_n\: F_n $$
Where 
$$ C_n = \frac{B_{2n} (-4)^n \left(1-4^n\right)\left(2n - 1\right)}{(2n)!} ,\qquad F_n = \int_{0}^{\frac{\pi}{4}} \frac{u^{2n - 2}}{u + 1} \: du$$
Taking the Taylor expansion for $\frac{1}{1 + u}$ at $u = 0$ (as given here) we can evaluate $F_n$:
\begin{align}
F_n &= \int_{0}^{\frac{\pi}{4}} \frac{u^{2n - 2}}{u + 1} \: du \\
&= \left[\ln|u + 1| + \sum_{k = 1}^{2n - 2} \left(-1\right)^k u^k \right]_{0}^{\frac{\pi}{4}} \\
&= \ln\left|\frac{\pi}{4} + 1 \right| + \sum_{k = 1}^{2n - 2} \left(-1\right)^k \frac{\pi^k}{4^kk}
\end{align}
From which we arrive at:
\begin{align}
I =  \sum_{n = 1}^{\infty}  C_n\:F_n &=  \sum_{n = 1}^{\infty}  C_n \left[\ln\left|\frac{\pi}{4} + 1 \right| + \sum_{k = 1}^{2n - 2} \left(-1\right)^k \frac{\pi^k}{4^kk} \right] \\
&= \ln\left|\frac{\pi}{4} + 1 \right| \sum_{n = 1}^{\infty}  C_n + \sum_{n = 1}^{\infty}\sum_{k = 1}^{2n - 2} C_n \left(-1\right)^k \frac{\pi^k}{4^kk}
\end{align}
Recall that
$$ \tan^{2}(u) + 1 = \sum^{\infty}_{n=1} \frac{B_{2n} (-4)^n \left(1-4^n\right)\left(2n - 1\right)}{(2n)!} u^{2n-2} = \sum_{n = 1}^{\infty} C_n u^{2n - 2} $$
Hence, 
$$ \tan^{2}(1) + 1 = \sec^{2}(1) = \sum_{n = 1}^{\infty} C_n$$
Thus, 
\begin{align}
 I &= \ln\left|\frac{\pi}{4} + 1 \right| \sum_{n = 1}^{\infty}  C_n + \sum_{n = 1}^{\infty}\sum_{k = 1}^{2n - 2} C_n \left(-1\right)^k \frac{\pi^k}{4^kk} \\
&= \ln\left|\frac{\pi}{4} + 1 \right|\sec^2(1) + \sum_{n = 1}^{\infty}\sum_{k = 1}^{2n -2}\frac{\left(-1\right)^kB_{2n} (-4)^n \left(1-4^n\right)\left(2n - 1\right)\pi^k}{(2n)!4^kk} 
\end{align}
However at this stage, I'm lost as to how this could be simplified. 
A: Let's take a look at that big 'ole fraction. 
$$Q_{nk}=\frac{\left(-1\right)^kB_{2n} (-4)^n \left(1-4^n\right)\left(2n - 1\right)\pi^k}{(2n)!4^kk}$$
$$Q_{nk}=\frac{\left(-1\right)^kB_{2n} (-1)^n 4^n\left(1-4^n\right)\left(2n - 1\right)\pi^k}{(2n)!4^kk}$$
$$Q_{nk}=\frac{\left(-1\right)^{n+k}B_{2n} 4^{n-k}\left(1-4^n\right)\left(2n - 1\right)\pi^k}{(2n)!k}$$
$$Q_{nk}=\frac{\left(-1\right)^{n+k}B_{2n} 4^{n-k}\left(1-4^n\right)\left(2n - 1\right)\pi^k}{(2n)(2n-1)(2n-2)!k}$$
$$Q_{nk}=\frac{\left(-1\right)^{n+k}B_{2n} 4^{n-k}\left(1-4^n\right)\pi^k}{(2nk)(2n-2)!}$$
Which is the tiniest bit more compact, but still probably isn't enough.
Edit: here's a little more
$$Q_{nk}=\frac{\left(-1\right)^{n+k}B_{2n} 2^{2(n-k)}\left(1-4^n\right)\pi^k}{(2nk)(2n-2)!}$$
$$Q_{nk}=\frac{\left(-1\right)^{n+k}B_{2n} 2^{2n-2k-1}\left(1-4^n\right)\pi^k}{nk(2n-2)!}$$
A: Starting from
$$ \cos(x) = \prod_{n\geq 0}\left(1-\frac{4x^2}{(2n+1)^2 \pi^2}\right) \tag{1}$$
and applying $\frac{d^2}{dx^2}\log(\cdot)$ to both sides we get
$$ \frac{1}{\cos^2 x} = 8\sum_{n\geq 0}\frac{\pi^2(2n+1)^2+4x^2}{(\pi^2(2n+1)^2-4x^2)^2}\tag{2}$$
By applying $\int_{0}^{\pi/4}\left(\cdot\right)\frac{dx}{x+1}$ we get that $I$ can be represented as a (horrible) series whose general term is $O\left(\frac{\log n}{n^2}\right)$. A numerical approximation of the given integral is better performed through some version of the Shafer-Fink inequality, but the series representation gives that there is a (feeble) relation between the given integral and $\zeta'(2)$, which on its turn is related to the Glaisher-Kinkelin constant. Shafer-Fink in its original formulation leads to
$$ \int_{0}^{1}\frac{dx}{1+\arctan x}\approx \int_{0}^{1}\frac{1+2\sqrt{1+x^2}}{1+3x+2\sqrt{1+x^2}}\,dx=\int_{1}^{1+\sqrt{2}}\frac{(1+t^2)(1+t+t^2)}{t^2(5t^2+2t-1)}\,dt $$
where the RHS can be computed by partial fraction decomposition. It equals $\approx \color{green}{0.714}\color{red}{504} $.
