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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.

StubbornAtom
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Renascence_5.
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  • Its numerical value is about $1.5301321254055419047226345952830\ldots$ – Lucian Jan 27 '16 at 14:03
  • @Lucian I know,but how to evaluate the closed form took me a long time and I still don't know how to do it. – Renascence_5. Jan 27 '16 at 15:07
  • I am not sure if the following may be useful, but WolframAlpha does have an exact antiderivative (http://bit.ly/1JFQ2hr) for the summand. For x=0, the antiderivative evaluates to zero, so the summand proves to be the antiderivative of the function evaluated at 1/n. I am not sure if this can be summed using standard techniques / simplifies the summation. – KR136 Jan 27 '16 at 21:26
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    Using the integral representation of the Zeta function http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/Zeta/07/01/01/, you can recast your sum in the integral form $$ \int_0^\infty dt\ \frac{4 \sqrt{t} , _1F_2\left(1;\frac{5}{4},\frac{7}{4};-\frac{t^2}{4}\right)}{\left(3 \sqrt{\pi }\right) (\mathrm{e}^t-1)} $$, which does not seem to have a simple closed form either...but may be simpler to analyze – Pierpaolo Vivo Jan 27 '16 at 22:05
  • @PierpaoloVivo, this integral seems to heavily dominated by the region around $0_+$, this should in principle be good enough to find a nice approximation. – tired Jan 28 '16 at 00:14
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    $=\sum_{n=1}^{\infty }\int_{0}^{\frac{1}{n}}\sqrt{x}\sum_{k=0}^{\infty}(-1)^kx^{2k}\mathrm{d}x=\sum_k(-1)^k\sum_n\int_0^{1/n}x^{2k+1/2}\mathrm{d}x$ yiels the $\zeta$-function formula. – A.S. Jan 28 '16 at 20:28
  • $$\zeta(2k+3/2) = \int_{0}^{+\infty}\frac{x^{2k+1/2}}{(2k+1/2)!}\cdot\frac{dx}{e^x-1},dx $$ gives $$ \sum_{n\geq 1}\int_{0}^{1/n}\frac{\sqrt{x}}{x^2+1},dx = \int_{0}^{+\infty}\frac{\sqrt{x}}{e^x-1}\underbrace{\sum_{k\geq 0}\frac{(-1)^k x^{2k}}{(2k+3/2)!}}{f(x)} ,dx$$ where $$ f(x) = \frac{g(x)+g(-x)}{2},\qquad g(x)=\sum{m\geq 0}\frac{i^m x^{m}}{(m+3/2)!}=\frac{1}{x^{2}}\left(\frac{2ix}{\sqrt{\pi}}-e^{ix}\sqrt{i x},\text{Erf}(\sqrt{ix})\right) $$

    $$ f(x) = -\frac{1}{x^2}\text{Re}\left(e^{ix}\sqrt{ix},\text{Erf}(\sqrt{ix})\right) $$

     – Jack D'Aurizio Jun 28 '20 at 22:02

1 Answers1

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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:
$$ I(n) = \dfrac{\sqrt n}{n+1}\left(1+\dfrac13\dfrac{2n}{(n+1)^2}+\dfrac15\dfrac{(2n)^2}{(n+1)^4}+\dots++\dfrac1{2k+1}\dfrac{(2n)^k}{(n+1)^{2k}}+\dots\right)$$$$+\dfrac12\,\dfrac{\sqrt n}{n-1}\left(1-\dfrac13\dfrac{2n}{(n-1)^2}+\dfrac15\dfrac{(2n)^2}{(n-1)^4}-\dots+\dfrac1{2k+1}\dfrac{(-2n)^k}{(n-1)^{2k}}+\dots\right), $$ $$I(1) = \dfrac{\pi}{4\sqrt2}.$$

Yuri Negometyanov
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