How to prove this?!🤔

\[\displaystyle\int_\tfrac{1}{2}^1\dfrac{\psi(x)}{1+\Gamma^2(x)}dx=\ln\left(\sqrt{\dfrac{1+\pi}{2\pi}}\right)\]

where \(\Gamma(x)\) and \(\psi(x)\) are the gamma and digamma functions respectively.

#Integral #definiteintegral #gammafunction #digammafunction #pi #logarithm

SOME EXOTIC INTEGRALS:

\[\displaystyle\int_0^1\dfrac{dx}{x^x}=\sum_{k=1}^\infty\dfrac{1}{k^k}\]
\[\displaystyle\int_0^1x^x\ dx=\sum_{k=1}^\infty\dfrac{(-1)^{k-1}}{k^k}\]
\[\displaystyle\int_0^1x^{x^2}\ dx=\sum_{k=1}^\infty\dfrac{(-1)^{k-1}}{(2k-1)^k}\]
\[\displaystyle\int_0^1x^{\sqrt{x}}\ dx=\sum_{k=1}^\infty(-1)^{k-1}\left(\dfrac{2}{k+1}\right)^k\]

\[\boxed{\boxed{\displaystyle\int_0^1x^{cx^a}\ dx=\sum_{k=1}^\infty\dfrac{(-c)^{k-1}}{((k-1)a+1)^k}}}\]
#integral #calculus #definiteintegral

SOME EXOTIC INTEGRALS:

\[\displaystyle\int_0^1\dfrac{dx}{x^x}=\sum_{k=1}^\infty\dfrac{1}{k^k}\]
\[\displaystyle\int_0^1x^x\ dx=\sum_{k=1}^\infty\dfrac{(-1)^{k-1}}{k^k}\]
\[\displaystyle\int_0^1x^{x^2}\ dx=\sum_{k=1}^\infty\dfrac{(-1)^{k-1}}{(2k-1)^k}\]
\[\displaystyle\int_0^1x^{\sqrt{x}}\ dx=\sum_{k=1}^\infty(-1)^{k-1}\left(\dfrac{2}{k+1}\right)^k\]

\[\boxed{\displaystyle\int_0^1x^{cx^a}\ dx=\sum_{k=1}^\infty\dfrac{(-c)^{k-1}}{((k-1)a+1)^k}}\]
#integral #calculus #definiteintegral