Integral involving hypergeometric function

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#1 h May 27, 2018, 9:30 AM • 1 Y

Y by Adventure10

Greetings, show that $\int_{0}^{1} \, _2F_1\left(\frac{1} {2}, 1;\frac{3} {2} ;-x^2 \right)\,dx=G$ Where G is the Catalan constant: https://en.wikipedia.org/wiki/Catalan%27s_constant

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#2 h May 27, 2018, 9:02 PM • 2 Y

Y by Adventure10, Mango247

we know that ${}_{2}{{F}_{1}}\left( a,b;c;z \right)=\sum\limits_{n=0}^{+\infty }{\frac{{{\left( a \right)}_{n}}{{\left( b \right)}_{n}}}{{{\left( c \right)}_{n}}}\cdot \frac{{{z}^{n}}}{n!}}$ for $\left| z \right|<1$
where ${{\left( q \right)}_{n}}=\left\{ \begin{matrix} 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n=0 \\ q\left( q+1 \right)\cdot \cdot \cdot \left( q+n-1 \right)\ \ \ n>0 \\ \end{matrix} \right.$

then
$\int\limits_{0}^{1}{{}_{2}{{F}_{1}}\left( \frac{1}{2},1;\frac{3}{2};-{{x}^{2}} \right)dx}=\int\limits_{0}^{1}{\sum\limits_{n=0}^{+\infty }{\frac{{{\left( \frac{1}{2} \right)}_{n}}\cdot {{\left( 1 \right)}_{n}}}{{{\left( \frac{3}{2} \right)}_{n}}}\cdot \frac{{{\left( -{{x}^{2}} \right)}^{n}}}{n!}dx}}=\int\limits_{0}^{1}{\left( 1+\sum\limits_{n=1}^{+\infty }{\frac{{{\left( \frac{1}{2} \right)}_{n}}\cdot {{\left( 1 \right)}_{n}}}{{{\left( \frac{3}{2} \right)}_{n}}}\cdot \frac{{{\left( -{{x}^{2}} \right)}^{n}}}{n!}} \right)dx}$

$=\int\limits_{0}^{1}{\left( 1+\sum\limits_{n=1}^{+\infty }{\frac{\left( \frac{1}{2}\left( \frac{1}{2}+1 \right)\left( \frac{1}{2}+2 \right)\cdot \cdot \cdot \left( \frac{1}{2}+n-1 \right) \right)\cdot \left( 1\cdot 2\cdot 3\cdot \cdot \cdot \left( n-1 \right)n \right)}{\left( \frac{3}{2}\left( \frac{3}{2}+1 \right)\left( \frac{3}{2}+2 \right)\cdot \cdot \cdot \left( \frac{3}{2}+n-1 \right) \right)}\cdot \frac{{{\left( -{{x}^{2}} \right)}^{n}}}{n!}} \right)dx}$

$=\int\limits_{0}^{1}{\left( 1+\sum\limits_{n=1}^{+\infty }{\frac{\frac{1\cdot 3\cdot 5\cdot \cdot \cdot \left( 2n-1 \right)}{{{2}^{n-1}}}}{\frac{3\cdot 5\cdot 7\cdot \cdot \cdot \left( 2n+1 \right)}{{{2}^{n-1}}}}\cdot {{\left( -{{x}^{2}} \right)}^{n}}} \right)dx}=\int\limits_{0}^{1}{\left( 1+\sum\limits_{n=1}^{+\infty }{\frac{{{\left( -{{x}^{2}} \right)}^{n}}}{2n+1}} \right)dx}=\int\limits_{0}^{1}{\left( \sum\limits_{n=0}^{+\infty }{\frac{{{\left( -{{x}^{2}} \right)}^{n}}}{2n+1}} \right)dx}=\sum\limits_{n=0}^{+\infty }{\frac{{{\left( -1 \right)}^{n}}}{{{\left( 2n+1 \right)}^{2}}}}=G$

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