Question

انتگرال معین زیر را بیابید: $ \int _0^1 Li_2( \sqrt{x}) $ از تعریف تابع$Li_2$ و تابع زتای ریمان حاصل به صورت $ \xi (2)- \frac{1}{2} H_2= \frac{ \pi ^{2} }{6} - \frac{3}{4} $ است.مراحل حل مد نظرم است.

Answer 1

$$we know,Li_a(z)= \sum _1^ \infty \frac{ z^{n} }{ n^{a} } so we can write,Li_2( \sqrt{x} )= \sum _1^ \infty \frac{ \sqrt{x} ^{n} }{ n^{2} } so,I= \int _0^ 1 \sum _1^ \infty \frac{ \sqrt{x} ^{n} }{ n^{2} } dx = \int _0^1 \sum _1^ \infty \frac{ x^{ \frac{n}{2} } }{ n^{2} } dx=\sum _1^ \infty \frac{1}{ n^{2} } \int _0^1 x^{ \frac{n}{2} } dx= \sum _1^ \infty \frac{1}{ n^{2} } ( \frac{ x^{ \frac{n}{2} +1} }{ \frac{n}{2}+1 } )= \sum _1^ \infty \frac{1}{ n^{2} } \frac{1}{( \frac{n}{2} +1)} ;; \frac{1}{ n^{2} ( \frac{n}{2} +1)} = \frac{-1}{2n} + \frac{1}{ n^{2} }+ \frac{1}{4} ( \frac{1}{ \frac{n}{2}+1 } )= \frac{1}{ n^{2} } - \frac{1}{2}( \frac{1}{n} )+ \frac{1}{2}( \frac{1}{n+2} ) \Longrightarrow \sum _1^ \infty \frac{1}{ n^{2}( \frac{n}{2} +1)}= \sum _1^ \infty \frac{1}{ n^{2} }- \frac{1}{2} \sum _1^ \infty ( \frac{1}{n} - \frac{1}{n-2} )= \zeta (2)- \frac{1}{2}H_2= \frac{ \pi ^{2} }{6}- \frac{1}{2}(1+ \frac{1}{2} ) = \frac{ \pi ^{2} }{6} - \frac{3}{4} $$