Question

نشان دهید: $$I(n)= \int _0^1 \frac{lnxln(1+ x^{ 2^{n} } )}{x} dx= \frac{-3 \zeta (3)}{ 2^{2n+2} } $$

Answer 1

$$ \leadsto u= x^{ 2^{n} } \Longrightarrow I= \int_0^1 \frac{lnu}{ 2^{n} }.ln(1+u). \frac{1}{ 2^{n} }. u^{ \frac{1}{ 2^{n} } -1}. \frac{du}{ u^{ \frac{1}{2n} } }= \frac{-1}{ 2^{2 ^{n} } } \sum_ {n=1} ^ \infty \frac{ (-1)^{n+1} }{ n^{3} }=- \frac{1}{ 2^{2n} } ( \frac{7}{8} \zeta (3)- \frac{1}{8} \zeta (3))$$

Comments

ادامه دارد.