Find $\sum_{n=1}^{\infty}\int_0^{\frac{1}{\sqrt{n}}}\frac{2x^2}{1+x^4}dx$

Solution 1:

You can simplify your expression using the two following identities

ArcTan[a] + ArcTan[b] = ArcTan[(a + b) / (1- a b)]

Log[(1 - a) / (1 + a)] = 2 ArcTanh[a]

Solution 2:

A more manageable way should be the following:

Let us make the change of the variable: $x=\frac{y}{\sqrt{n}}$

$$a_n=\int_0^{\frac{1}{\sqrt{n}}}\frac{2x^2}{1+x^4}dx=\frac{2}{n^{\frac{3}{2}}}\int_0^1\frac{y^2}{1+\frac{y^4}{n^2}}dy$$

Now, let's expand the integrand:

$$\frac{y^2}{1+\frac{y^4}{n^2}}=\sum_{k=0}^\infty(-1)^k\frac{y^{4k+2}}{n^{2k}}$$ and

$$\int_0^1\frac{y^2}{1+\frac{y^4}{n^2}}dy=\sum_{k=0}^\infty\frac{(-1)^k}{4k+3}\frac{1}{n^{2k}}$$ and

$$a_n=\int_0^{\frac{1}{\sqrt{n}}}\frac{2x^2}{1+x^4}dx=2\sum_{k=0}^\infty\frac{(-1)^k}{4k+3}\frac{1}{n^{2k+\frac{3}{2}}}$$

Finally, the sum:

$$S=\sum_{n=1}^\infty a_n=2\sum_{k=0}^\infty\frac{(-1)^k}{4k+3}\sum_{n=1}^\infty\frac{1}{n^{2k+\frac{3}{2}}}=$$

$$=2\sum_{k=0}^\infty\frac{(-1)^k}{4k+3}\zeta(2k+\frac{3}{2})$$

where $\zeta(s)=\sum_{k=1}^\infty\frac{1}{k^s}$ is the Riemann zeta function.