Question

$\red{ \displaystyle \rm\int_{0}^{1} \frac{ {x}^{ \frac{1}{a} } - 1}{ \left(1 + {x}^{ \frac{1}{a} }\right) ln(x) } dx}$​

Answer 1

$\red{ \displaystyle \rm\int_{0}^{1} \frac{ {x}^{ \frac{1}{a} } - 1}{ \left(1 + {x}^{ \frac{1}{a} }\right) ln(x) } dx}$

$\red{ \displaystyle \rm\int_{0}^{1} \frac{ y - 1}{ \left(1 + y^{ }\right) \: a \ln(y) } \: a {y}^{a - 1} \: dy } \:$

$\red{ \displaystyle \rm\int_{0}^{1} \frac{ {y}^{a} - {y}^{a - 1} }{ \left(1 + y^{ }\right) \ln(y) } \: dy } \:$

$\red{ \displaystyle \rm\int_{0}^{1} \frac{ {y}^{a - 1} - {y}^{a } }{ 1 - {y}^{2} } \overbrace{\left [ - \frac{1 - y}{ ln(y) } \right] } ^{ \displaystyle\int_{0}^1 \rm{y}^t \: dt } \: dy } \:$

$\red{ \displaystyle \rm\int_{0}^{1} \int_{0}^{1} \frac{ {y}^{t + a - 1} - {y}^{t + a} }{1 - {y}^{2} } \: dy \: dt }$

$\red{ \displaystyle \rm \frac{1}{2} \int_{0}^{1} \int_{0}^{1} \frac{ {y}^{ \frac{t}{2} + \frac{a}{2} - 1 } - {y}^{ \frac{t}{2} + \frac{a}{2} - \frac{1}{2} \frac{}{}}}{1 - {y}^{} } \: dy \: dt }$

$\red{ \displaystyle \rm \frac{1}{2} \int_{0}^{1} \left[\int_{0}^{1} \frac{1 - {y}^{ \frac{t}{2} + \frac{a}{2} - \frac{1}{2} } {}}{1 - {y}^{} } \: dy \: -\int_{0}^{1} \frac{1 - {y}^{ \frac{t}{2} + \frac{a}{2} - 1 } {}}{1 - {y}^{} } \: dy \right ] \: dt }$

$\red{ \displaystyle \rm \frac{1}{2} \int_{0}^{1} \int_{0}^{1} \left[ \Psi \left(\frac{t}{2} + \frac{a}{2} + \frac{1}{2} \right)- \Psi \left(\frac{t}{2} + \frac{a}{2} \right) \: \right] \: dt }$

$\rm \red{ ln \left( \frac{\Gamma( \frac{t}{2} + \frac{a}{2} + \frac{1}{2}) }{\Gamma( \frac{t}{2} + \frac{a}{2}) } \right)^{t = 1}_{t = 0}}$

$\rm \red{ ln \left( \frac{\Gamma( \frac{a}{2} + 1) }{\Gamma( \frac{1}{2} + \frac{a}{2}) } \right) - ln \left( \frac{\Gamma( \frac{a}{2} + \frac{1}{2} ) }{\Gamma( \frac{a}{2}) } \right) }$

$\rm \red{ ln \left( \frac{\Gamma( \frac{a}{2} + 1) \Gamma( \frac{a}{2}) }{\Gamma^{2} ( \frac{a}{2} + \frac{1}{2} ) } \right) }$