Question

$\large\displaystyle \bf\red {\int_{0}^{ \frac{1}{2} } \frac{ \Gamma(1 + x) \Gamma(1 - x)}{ \Gamma( \frac{1}{2} + x) \Gamma( \frac{1}{2} - x) } \: dx}$​

Answer 1

Answer:

Step-by-step explanation:

We know that,

$1)~~\Gamma({n+1}) = {n}\Gamma{n}\\2)~~\Gamma(n).\Gamma(1-n) = \frac{\pi}{sin{n\pi}} , 0 < n < 1 (\text{Duplication formula})$

Now,

$\Gamma(1+x)\cdot \Gamma(1-x)\\ = x\Gamma{x}\cdot \Gamma(1-x)\\ = x\cdot \frac{\pi}{sin{x\pi}}\quad (~\because~0 < x < \frac{1}{2}~)$

$\text{and}~~ \Gamma(\frac{1}{2} +x)\cdot \Gamma(\frac{1}{2} - x)\\= \Gamma(\frac{1}{2} +x)\cdot \Gamma(1-(\frac{1}{2} + x))\\= \frac{\pi}{sin{(\frac{1}{2}+x)\pi}}\\= \frac{\pi}{sin{(\frac{\pi}{2}+x\pi)}}\\= \frac{\pi}{cos{x\pi}}$

Therefore,

$\large\displaystyle \bf\red {\int_{0}^{ \frac{1}{2} } \frac{ \Gamma(1 + x) \Gamma(1 - x)}{\Gamma(\frac{1}{2}+x) \Gamma(\frac{1}{2}-x)} dx$

$= \int\limits^\frac{1}{2}_ 0 \frac{x\cdot \frac{\pi}{sin{x\pi}}}{\frac{\pi}{cos{x\pi}}} \, dx$

$= \int\limits^\frac{1}{2}_0 {x cot{x\pi}} \,dx$

$= \frac{\log(2)}{2\pi}$

Answer 2

$\begin{gathered} \sf \red{1)~~\Gamma({n+1}) = {n}\Gamma{n}}\\ \sf \red{2)~~\Gamma(n).\Gamma(1-n) = \frac{\pi}{sin{n\pi}} , 0 < n < 1 (\text{Duplication formula})}\end{gathered}$

Now,

$\begin{gathered}\sf \red {\Gamma(1+x)\cdot \Gamma(1-x)}\\ \sf \red{= x\Gamma{x}\cdot \Gamma(1-x)}\\ \sf \red{= x\cdot \frac{\pi}{sin{x\pi}}\quad (~\because~0 < x < \frac{1}{2}~)}\end{gathered}$

$\begin{gathered}\sf \red{\text{and}~~ \Gamma(\frac{1}{2} +x)\cdot \Gamma(\frac{1}{2} - x)}\\ \sf \red{= \Gamma(\frac{1}{2} +x)\cdot \Gamma(1-(\frac{1}{2} + x))}\\ \sf \red{= \frac{\pi}{sin{(\frac{1}{2}+x)\pi}}}\\ \sf \red{= \frac{\pi}{sin{(\frac{\pi}{2}+x\pi)}}} \\ \sf \red{= \frac{\pi}{cos{x\pi}}}\end{gathered}$

Therefore,

$\large\displaystyle \bf\red {\int_{0}^{ \frac{1}{2} } \frac{ \Gamma(1 + x) \Gamma(1 - x)}{\Gamma(\frac{1}{2}+x) \Gamma(\frac{1}{2}-x)} dx}$

$\displaystyle \sf \red{= \int\limits^\frac{1}{2}_ 0 \frac{x\cdot \frac{\pi}{sin{x\pi}}}{\frac{\pi}{cos{x\pi}}} \, dx}$

$\displaystyle \sf \red{= \int\limits^\frac{1}{2}_0 {x cot{x\pi}} \,dx}$

$\sf \red{= \frac{\log(2)}{2\pi}}$