Question

نشان دهید که: $$ \int _0^ \frac{ \pi }{2}[log(tan x)+ log^{2}(tan x)] sin^{3} (2x)dx= \frac{ \pi ^{2} }{18} - \frac{1}{3} $$

Answer 1

$$I(k)= \int _0^ \frac{ \pi }{2} tan^{k} (x) sin^{3} (x)dx= 2^{3} \int _0^ \frac{ \pi }{3} \frac{ sin ^{k} (x)}{ cos ^{k} (x)} . sin ^{3} (x) cos ^{3} (x)dx= 2^{3} \int _0^ \frac{ \pi }{2} sin^{2(2+ \frac{k}{2} )-1} (x) cos^{2(2- \frac{k}{2} )-1} (x)dx \Longrightarrow I(k)=4 \beta (2+ \frac{k}{2} ,2- \frac{k}{2} )=4 \frac{ \Gamma (2+ \frac{k}{2} ) \Gamma (2- \frac{k}{2} )}{ \Gamma (4)} \Longrightarrow I(k)= \frac{2}{3} \Gamma (2+ \frac{k}{2} ) \Gamma (2- \frac{k}{2} ) \wedge 2 \int _0^ \frac{ \pi }{2} sin^{2m-1} (x) cos^{2n-1}(x) dx= \beta (m,n) \Longrightarrow I= \frac{dI(k)}{dk} + \frac{ d^{2} I(k)}{d k^{2} } | _ {k=0} = \frac{2}{3} [ \frac{1}{4} 2 \Gamma ^{2} (2) \psi '(2)]= \frac{1}{3} \psi '(2)= \frac{ x^{k} }{8} - \frac{1}{3} $$ $$Note: \frac{dI(k)}{dk} = \frac{1}{2} [ \psi (2+ \frac{k}{2} )- \psi (2- \frac{k}{2} )] \Gamma (2+ \frac{k}{2} ) \Gamma (2- \frac{k}{2} ) at k=0 \wedge \frac{dI(k)}{dk} =0 \wedge \frac{ d^{2} I(k)}{dk} = \Gamma (2+ \frac{k}{2} ) \Gamma (2- \frac{k}{2} ) [ \psi (2+ \frac{k}{2} )- \psi (2- \frac{k}{2} )]^{2} +[ \psi '(2+ \frac{k}{2} )+ \psi '(2- \frac{k}{2} )] at k=0 \wedge \frac{ d^{2} I(k)}{dk} =2 \Gamma ^{2} \psi '(2)$$