3
$\begingroup$

Context: this is a purely recreational problem that I came up with myself.

I am trying to extend this colsed form:

$$\sum_{n\geq1}\frac{x^n}{n{2n\choose n}}=2\sqrt{\frac{x}{4-x}}\arctan\sqrt{\frac{x}{4-x}},\qquad 0<x<4$$ I know that $S(x)=\sum_{n\geq1}\frac{x^n}{n{2n\choose n}}$ converges for $|x|<4$, but the closed form above only works for half that range. Is there a closed form for the other half?

I think that it might come down to figuring out how to integrate $$\int_0^1\frac{\mathrm dt}{xt^2-xt+1}$$ when $-4<x<0$, but I'm not sure. In your answer, please show your proof. Thanks.


For those interested, a proof.

Noting that ${2n\choose n}^{-1}=\frac{n\Gamma^2(n)}{2\Gamma(2n)}$, $$S(x)=\frac12\sum_{n\geq1}\int_0^1x^n[t(1-t)]^{n-1}\mathrm dt$$ Then using the geometric series, $$S(x)=\frac{x}2\int_0^1\frac{\mathrm dt}{xt^2-xt+1}$$ Then we focus on $$I(k;a,b,c)=\int_0^k\frac{\mathrm dt}{at^2+bt+c}$$ If we complete the square in the denominator then preform the appropriate trig substitution, we see that for $4ac>b^2$, $$I(k;a,b,c)=\frac2{\sqrt{4ac-b^2}}\bigg[\arctan\frac{2ak+b}{\sqrt{4ac-b^2}}-\arctan\frac{b}{\sqrt{4ac-b^2}}\bigg]$$ So for $0<x<4$, $$S(x)=\frac{x}2I(1;x,-x,1)$$ $$S(x)=2\sqrt{\frac{x}{4-x}}\arctan\sqrt{\frac{x}{4-x}}$$ As desired.

$\endgroup$
  • 3
    $\begingroup$ $-i\arctan(ix)=\text{arctanh}(x)=\frac{1}{2}\log\frac{1+x}{1-x}$ allows to cover the whole $(-4,4)$. $\endgroup$ – Jack D'Aurizio Jan 11 at 17:25
  • 2
    $\begingroup$ You are actually just integrating the Maclaurin series of the squared arcsine. $\endgroup$ – Jack D'Aurizio Jan 11 at 17:26
  • 1
    $\begingroup$ The integral you asked for equals $\frac{4 \tanh ^{-1}\left(\sqrt{\frac{x}{x-4}}\right)}{\sqrt{(x-4) x}}$ $\endgroup$ – ablmf Jan 11 at 21:23
  • $\begingroup$ @ablmf how do you find this? $\endgroup$ – clathratus Jan 11 at 21:46
  • $\begingroup$ @clathratus Mathematica. BTW, the anti-derivative is $\frac{2 \tan ^{-1}\left(\frac{(2 t-1) \sqrt{x}}{\sqrt{4-x}}\right)}{\sqrt{-(x-4) x}}$. So this is the proof. $\endgroup$ – ablmf Jan 11 at 21:51

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.