انتگرال معین زیر را بیابید.

Question

مطلوب است محاسبه انتگرال معین زیر: $$ \int _0^ \frac{ \pi }{2} \sqrt{ \frac{ sin^{3} (\theta) }{ cos ^{7}( \theta) } } \frac{ln(tan (\theta))}{ e^{tan (\theta) } + e^{-tan( \theta) } } d \theta=?$$

Comments

$$I= \int _0^ \frac{ \pi }{2}  \sqrt{ tan^{3}  \theta }  \frac{lntan \theta }{ e^{tan \theta } + e^{-tan \theta } }  sec^{2}  \theta d \theta \wedge tan  \theta =x \Longrightarrow I= \int _0^ \infty  \frac{ x^{ \frac{3}{2} } lnx}{ e^{x} + e^{-x} } dx \Longrightarrow I(k)= \int _0^ \infty  \frac{ x^{k} }{ e^{x} + e^{-x} } dx= \int _0^ \infty  \frac{ e^{-x}  x^{k} }{1+ e^{-2x} } dx= \int _0^ \infty  x^{k}  \sum _{n=0} ^ \infty  (-1)^{n}  e^{-2nx-x} dx= \sum _ {n=0} ^ \infty  (-1)^{n}  \int _0^ \infty  e^{-x(2n+1)}  x^{k} dx= \Gamma (k+1) \sum _ {n=0} ^ \infty  \frac{ (-1)^{n} }{ (2n+1)^{2} } = \Gamma (k+1) \beta (k+1)$$$$ \frac{\partial}{\partial x} I(k)= \frac{\partial}{\partial x}  \int _0^ \infty  \frac{ x^{k} }{ e^{x} + e^{-x} } dx= \int _0^ \infty  \frac{ x^{k} lnx}{ e^{x} + e^{-x} } dx \Longrightarrow I'(k)= \int _0^ \infty  \frac{ x^{k} lnx}{ e^{x} + e^{-x} } dx= \beta '(k+1) \Gamma (k+1)+ \beta (k+1) \Gamma (k+1) \psi (k+1) \Longrightarrow I'( \frac{3}{2} )=I= \beta '( \frac{5}{2} ) \Gamma ( \frac{5}{2} )+ \beta ( \frac{5}{2} ) \Gamma ( \frac{5}{2} ) \psi ( \frac{5}{2} ) \wedge  \psi (m+ \frac{1}{2} )=2 \psi (2m)- \psi (m)-2ln2 \Longrightarrow  \psi ( \frac{5}{2} )= \frac{8}{3} - \gamma -2ln2 \Longrightarrow I'= \frac{3 \sqrt{ \pi } }{4} [  \beta '( \frac{5}{2} )+[ \frac{8}{3} - \gamma -2ln2] \beta ( \frac{5}{2} )]$$

Answer 1

$$I= \int _0^ \frac{ \pi }{2} \sqrt{ tan^{3} \theta } \frac{lntan \theta }{ e^{tan \theta } + e^{-tan \theta } } sec^{2} \theta d \theta \wedge tan \theta =x \Longrightarrow I= \int _0^ \infty \frac{ x^{ \frac{3}{2} } lnx}{ e^{x} + e^{-x} } dx \Longrightarrow I(k)= \int _0^ \infty \frac{ x^{k} }{ e^{x} + e^{-x} } dx= \int _0^ \infty \frac{ e^{-x} x^{k} }{1+ e^{-2x} } dx= \int _0^ \infty x^{k} \sum _{n=0} ^ \infty (-1)^{n} e^{-2nx-x} dx= \sum _ {n=0} ^ \infty (-1)^{n} \int _0^ \infty e^{-x(2n+1)} x^{k} dx= \Gamma (k+1) \sum _ {n=0} ^ \infty \frac{ (-1)^{n} }{ (2n+1)^{2} } = \Gamma (k+1) \beta (k+1)$$$$ \frac{\partial}{\partial x} I(k)= \frac{\partial}{\partial x} \int _0^ \infty \frac{ x^{k} }{ e^{x} + e^{-x} } dx= \int _0^ \infty \frac{ x^{k} lnx}{ e^{x} + e^{-x} } dx \Longrightarrow I'(k)= \int _0^ \infty \frac{ x^{k} lnx}{ e^{x} + e^{-x} } dx= \beta '(k+1) \Gamma (k+1)+ \beta (k+1) \Gamma (k+1) \psi (k+1) \Longrightarrow I'( \frac{3}{2} )=I= \beta '( \frac{5}{2} ) \Gamma ( \frac{5}{2} )+ \beta ( \frac{5}{2} ) \Gamma ( \frac{5}{2} ) \psi ( \frac{5}{2} ) \wedge \psi (m+ \frac{1}{2} )=2 \psi (2m)- \psi (m)-2ln2 \Longrightarrow \psi ( \frac{5}{2} )= \frac{8}{3} - \gamma -2ln2 \Longrightarrow I'= \frac{3 \sqrt{ \pi } }{4} [ \beta '( \frac{5}{2} )+[ \frac{8}{3} - \gamma -2ln2] \beta ( \frac{5}{2} )]$$