<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] Really what you want to do is define the function 
$$f_n(x)=\int_0^x\frac{t^{2n}}{1+t^2}dt=\int_0^xt^{2n}\sum_{k\geq0}(-1)^kt^{2k}dt\\
=\sum_{k\geq0}\frac{(-1)^k}{2n+2k+1}x^{2n+2k+1}$$
Then notice that the sum you seek is given by 
$$g(n)=\int_0^1 f_n(x)\frac{dx}{x}=\sum_{k\geq0}\frac{(-1)^k}{(2n+2k+1)^2}=\frac14\sum_{k\geq0}\frac{(-1)^k}{(\frac{2n+1}2+k)^2}$$
To compute this, we recall the definition
$$\psi_m(x)=\left(\frac{d}{dx}\right)^{m+1}\log\Gamma(x)=(-1)^{m+1}m!\sum_{k\geq0}\frac1{(x+k)^{m+1}}$$
So we have that 
$$\begin{align}
\sum_{k\geq0}\frac{(-1)^k}{(x+k)^m}&=\sum_{k\geq0}\frac{(-1)^{2k}}{(x+2k)^m}+\sum_{k\geq0}\frac{(-1)^{2k+1}}{(x+2k+1)^m}\\
&=2^{-m}\sum_{k\geq0}\frac1{(\frac{x}2+k)^m}-2^{-m}\sum_{k\geq0}\frac1{(\frac{x+1}2+k)^m}\\
&=\frac{(-1)^m}{2^m(m-1)!}\left[\psi_{m-1}\left(\frac{x}2\right)-\psi_{m-1}\left(\frac{x+1}2\right)\right]\tag{1}
\end{align}$$
So just plug in $m=2$ and $x=\frac{2n+1}{2}$ into $(1)$ to find $g(n)$:
$$\begin{align}
g(n)&=\frac14\sum_{k\geq0}\frac{(-1)^k}{(\frac{2n+1}2+k)^2}\\
&=\frac14\cdot\frac{(-1)^2}{2^2(1)!}\left[\psi_{1}\left(\frac{2n+1}4\right)-\psi_{1}\left(\frac{2n+3}4\right)\right]\\
&=\frac1{16}\left[\psi_{1}\left(\frac{2n+1}4\right)-\psi_{1}\left(\frac{2n+3}4\right)\right]
\end{align}$$
Alternatively, you could note that
$$f_n(x)=\int_0^x \frac{t^{2n}}{1+t^2}dt=x^{2n+1}\int_0^1 \frac{t^{2n}}{1+(xt)^2}dt$$
So you'd have the double integral
$$g(n)=\int_0^1\int_0^1\frac{(uv)^{2n}}{1+(uv)^2}dudv$$
Although evaluating such a double integral would require the use of the $\psi_1$ function, and would eventually lead us back to the result we just obtained.
</s>