The equation is $\int_{0}^{1}dy\left( \sqrt{1+\Pi ^{2}+2\Pi \sqrt{1-y^{2}}}-\sqrt{1+\Pi ^{2}-2\Pi \sqrt{1-y^{2}}}\right) =\frac{\pi \Pi }{2}\ _{2}F_{1}(-\frac{1}{2},% \frac{1}{2};2;\Pi ^{2})$, where $\Pi \le 1$. How to reduce the left integral into the right hypergeometric fucntion? Is there anyone can solve this problem? Thank you.

Thank you for @T. Amdeberhan. I learn general binomial expansion from your answer. Your result is more general. After some modifications, the result is \begin{eqnarray*} &&\int_{0}^{1}dy\left( \sqrt{1+\Pi ^{2}+2\Pi \sqrt{1-y^{2}}}-\sqrt{1+\Pi ^{2}-2\Pi \sqrt{1-y^{2}}}\right) \\ &=&\frac{\pi }{2}\frac{\Pi }{\sqrt{1+\Pi ^{2}}}\ _{2}F_{1}\left( \frac{1}{4},% \frac{3}{4};2;\left( \frac{2\Pi }{1+\Pi ^{2}}\right) ^{2}\right) , \end{eqnarray*} which is valid for any value of $\Pi $. But for $\Pi \leq 1$, how to transform the result into

$ \int_{0}^{1}dy\left( \sqrt{1+\Pi ^{2}+2\Pi \sqrt{1-y^{2}}}-\sqrt{1+\Pi ^{2}-2\Pi \sqrt{1-y^{2}}}\right) =\frac{\pi \Pi }{2}\ _{2}F_{1}(-\frac{1}{2},% \frac{1}{2};2;\Pi ^{2}) $?

It is seems that we should expand the integrad with respect to $\Pi $. Should we use multinomial expansion? But how to deal with the double summation? Could we prove the equality of the two hypergeometrical functions directly for $\Pi \leq 1$?

  • $\begingroup$ @T. Amdeberhan Could you check my updated question? $\endgroup$ – Rujiang Dec 12 '16 at 7:42

Letting $\beta=\frac{2\Pi}{1+\Pi^2}$, write the integrand as $$\sqrt{1+\Pi^2}\left(\sqrt{1+\beta \sqrt{1-y^{2}}}-\sqrt{1-\beta \sqrt{1-y^{2}}}\right).\tag1$$ Use the binomial expansion to express (1), after combining two infinite series, in the form $$2\sqrt{1+\Pi^2}\sum_{n\geq0}\binom{1/2}{2n+1}\beta^{2n+1}\int_0^1(1-y^2)^n\sqrt{1-y^2}dy.\tag2$$ Standard integral evaluations enable to compute the integral, hence (2) becomes $$\begin{align} &2\sqrt{1+\Pi^2}\sum_{n\geq0}\binom{1/2}{2n+1}\beta^{2n+1}\binom{2n+2}{n+1}\frac{\pi}{2^{2n+3}} \\ =& \frac{\beta\pi}4\sqrt{1+\Pi^2}\sum_{n\geq0}\binom{1/2}{2n+1}\binom{2n+2}{n+1}\left(\frac{\beta}2\right)^{2n} \\ =& \frac{\beta\pi}4\sqrt{1+\Pi^2}\sum_{n\geq0}\binom{4n}{2n}\frac1{(2n+1)2^{4n+1}}\binom{2n+2}{n+1}\left(\frac{\beta}2\right)^{2n} \\ =& \frac{\beta\pi}8\sqrt{1+\Pi^2}\sum_{n\geq0}\binom{4n}{2n}\frac1{2n+1}\binom{2n+2}{n+1}\left(\frac{\beta}8\right)^{2n} \\ =& \frac{\pi\Pi}{4\sqrt{1+\Pi^2}}\sum_{n\geq0}\binom{4n}{2n}\frac1{2n+1}\binom{2n+2}{n+1}\left(\frac{\beta}8\right)^{2n} \\ =& \frac{\pi\Pi}{4\sqrt{1+\Pi^2}}\sum_{n\geq0}\frac{(4n)!}{(2n)!(n+1)!n!}\left(\frac{\Pi}{4+4\Pi^2}\right)^{2n}. \end{align}$$

  • $\begingroup$ Thank you. Could you check my updated question? $\endgroup$ – Rujiang Dec 12 '16 at 7:42
  • 1
    $\begingroup$ The sum is more or less the definition of the hypergeometric function $_{2}F_{1}(1/4,3/4;2;4 \Pi^2/(1+\Pi^2)^2)$ (decorated with some factors). This is then to be transformed using, e.g., dlmf.nist.gov/15.8.E15 $\endgroup$ – Johannes Trost Dec 12 '16 at 12:27
  • $\begingroup$ This website is excellent! I have solved this problem. Thank you very much. @Johannes Trost $\endgroup$ – Rujiang Dec 12 '16 at 15:32

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