Evaluate $\sum_{n=1}^{\infty }\int_{0}^{\frac{1}{n}}\frac{\sqrt{x}}{1+x^{2}}\,\mathrm{d}x$

Question

I have some trouble in evaluating this series $$\sum_{n=1}^{\infty }\int_{0}^{\frac{1}{n}}\frac{\sqrt{x}}{1+x^{2}}\mathrm{d}x$$ I tried to calculate the integral first, but after that I found the series become so complicated. Besides, I found maybe the series equals to $$\sum_{k=0}^{\infty }\frac{\left ( -1 \right )^{k}\zeta \left ( 2k+\dfrac{3}{2} \right )}{2k+\dfrac{3}{2}}$$ this is definitely a monster to me. So I want to know is there a good way to solve this integral-series.

Answer 1

Let $$I(n) = \int_{0}^{\frac{1}{n}}\frac{\sqrt{x}}{1+x^{2}}\mathrm{d}x.$$ Substitution: $$x = t^2,\quad \mathrm{d}x = 2t\mathrm{d}t:$$ $$I(n) = \int\limits_{0}^{\frac{1}{\sqrt{n}}}\frac{2t^2}{1+t^{4}}\mathrm{d}t.$$ $$I(n) = \int\limits_{0}^{\frac{1}{\sqrt{n}}}\frac{2}{t^2 + \dfrac{1}{t^2}}\mathrm{d}t.$$ $$I(n) = \int\limits_{0}^{\frac{1}{\sqrt{n}}}\frac{1-\dfrac1{t^2}}{\left(t + \dfrac{1}{t}\right)^2-2}\mathrm{d}t + \int\limits_{0}^{\frac{1}{\sqrt{n}}}\frac{1+\dfrac1{t^2}}{\left(t - \dfrac{1}{t}\right)^2+2}\mathrm{d}t.$$ $$I(n) = \dfrac{1}{2\sqrt{2}}\left.\ln\left|\frac{t + \dfrac{1}{t}-\sqrt 2}{t + \dfrac{1}{t}+\sqrt 2}\right|\right|_0^{\frac1{\sqrt n}} + \dfrac1{2\sqrt2}\left.\arctan\dfrac{ {t - \dfrac{1}{t}}}{\sqrt2}\right|_0^{\frac1{\sqrt n}}.$$ $$I(n) = \dfrac{1}{2\sqrt{2}}\ln\frac{n - \sqrt {2n} + 1}{n + \sqrt {2n} + 1} + \dfrac1{2\sqrt{2}}\left(\dfrac{\pi}{2}-\arctan\dfrac{n - 1}{\sqrt{2n}}\right), \quad I(1) = \dfrac{\pi}{4\sqrt2}.$$ or $$I(n) = \dfrac{1}{2\sqrt{2}}\ln\frac{1 -\dfrac{\sqrt{2n}}{n + 1}}{1 +\dfrac{\sqrt{2n}}{n + 1}} + \dfrac1{2\sqrt{2}}\arctan\dfrac{\sqrt{2n}}{n - 1}, \quad I(1) = \dfrac{\pi}{4\sqrt2}.$$ UPD
Maclaurin series is as follows:
$$ I(n) = \dfrac{\sqrt n}{n+1}\left(1+\dfrac13\dfrac{2n}{(n+1)^2}+\dfrac15\dfrac{(2n)^2}{(n+1)^4}+\dots++\dfrac1{2k+1}\dfrac{(2n)^k}{(n+1)^{2k}}+\dots\right)$$$$+\dfrac12\,\dfrac{\sqrt n}{n-1}\left(1-\dfrac13\dfrac{2n}{(n-1)^2}+\dfrac15\dfrac{(2n)^2}{(n-1)^4}-\dots+\dfrac1{2k+1}\dfrac{(-2n)^k}{(n-1)^{2k}}+\dots\right), $$ $$I(1) = \dfrac{\pi}{4\sqrt2}.$$