<s>[INST] Having the identity: $\int_0^1 \frac{x^{2n}}{1+x^2} dx= \sum_{k=0}^{\infty} \frac{(-1)^k}{2n+2k+1}$, how do I square the fraction inside the sum? I have the integral:
\begin{align}
\int_0^1 \frac{x^{2n}}{1+x^2} dx &= \int_0^1 x^{2n} \sum_{k=0}^{\infty}(-1)^kx^{2k} dx\\
&=  \sum_{k=0}^{\infty} \frac{(-1)^k}{2n+2k+1}
\end{align}
For $n=0$ we have $\pi/4$. So, for some positive integer $n$, the integral gives something in terms of $\pi/4$.
I wish the fraction on right hand to have a squared denominator inside the sum, like this:
$$ \sum_{k=0}^{\infty} \frac{(-1)^k}{(2n+2k+1)^2} $$
Because for a positive integer $n$ it gives something in terms of the Catalan constant $(1-1/3^2+1/5^2-\cdots)$.
one way to do this is the following (due to Beukers):
\begin{align}
\int_0^1 \frac{x^{2n+y}}{1+x^2} dx &= \int_0^1x^{2n+y} \sum_{k=0}^{\infty}(-1)^k x^{2k} dx\\
&=  \sum_{k=0}^{\infty} \frac{(-1)^k}{2n+2k+1+y}
\end{align}
Differentiate inside the integral with respect to $y$:
\begin{align}
& \int_0^1 \frac{\partial}{\partial y} \frac{x^{2n+y}}{1+x^2} dx
= \sum_{k=0}^{\infty} \frac{\partial}{\partial y} \frac{(-1)^k}{n+k+1+y} \\
& \int_0^1 \frac{x^{2n+y}}{1+x^2} \ln x \: dx = \sum_{k=0}^{\infty} \frac{(-1)^{k+1}}{(n+k+1+y)^2}
\end{align}
for $y=0$
\begin{align}
\int_0^1  \frac{x^{2n}}{1+x^2} \ln x \: dx = \sum_{k=0}^{\infty} \frac{(-1)^{k+1}}{(n+k+1)^2}
\end{align}
The problem is the $\ln(x)$ because it makes the integral way too larger than before. If we chose some polynomial $(P_{2n}(x))$ with integer coefficients and even powers:
$$\int_0^1 P_{2n}(x)\ln (x)/(1+x^2)dx \gg \int_0^1 P_{2n}(x)/(1+x^2)dx$$
for some integer $n$. By the sign $\gg$ I mean that the integral is very "sensitivity" for values of $n$, even if the difference between the integrals are $0.01$ this already makes a big difference. 
Does anyone have another idea?
 [/INST] HINT: Manipulate this identity involving the Digamma Function $$\sum_{k=0}^\infty \frac{(-1)^k}{zk+1}= \frac{1}{2z}\left( \psi_0(\frac{1}{2}+\frac{1}{2z})-\psi_0(\frac{1}{2})\right)$$
to get $$\frac{1}{4}\left(\psi_0(\frac{n}{2}+\frac{3}{4})-\psi_0(\frac{n}{2}+\frac{1}{4})\right)$$ for the value of the integral.
</s>