نشان دهیدکه:: $$ \int _0^1 Li_{2} ( \sqrt{x})dx= \zeta (2)- \frac{1}{2} H_{2} = \frac{ \pi ^{2} }{6} - \frac{3}{4} $$

Question

نشان دهیدکه:

$$ \int _0^1Li_2 \sqrt{x} dx= \frac{ { \pi } ^{2} }{6} - \frac{3}{4} $$

$$ Li_{a} (z)= \sum _ {n=1} ^ \infty \frac{ z^{n} }{ n^{a} }$$

Answer 1

$$Li_2( \sqrt{x} )= \sum _ {n=1} ^ \infty \frac{ ( \sqrt{x} )^{n} }{ n^{2} } \Longrightarrow I= \int _0^1 \sum _ {n=1} ^ \infty \frac{ (\sqrt{x} )^{n} }{ n^{2} } dx= \int _0^1 \sum _ {n=1} ^ \infty \frac{ x^{ \frac{n}{2} } }{ n^{2} } dx= \sum _ {n=1} ^ \infty \frac{1}{ n^{2} } \int _0^1 x^{ \frac{n}{2} } dx= \sum _ {n=1} ^ \infty \frac{1}{ n^{2} } [ \frac{ x^{ \frac{n}{2} +1}}{ \frac{n}{2} +1}]= \sum _ {n=1} ^ \infty \frac{1}{ n^{2} } . \frac{1}{ \frac{n}{2}+1 } \Longrightarrow \frac{1}{ n^{2}( \frac{n}{2} +1)} = \frac{A}{n} + \frac{B}{ n^{2} } + \frac{C}{ \frac{n}{2} +1} \Longrightarrow B=1 \wedge C= \frac{1}{4} \wedge A=- \frac{1}{2} \Longrightarrow ?= \sum _ {n=1} ^ \infty \frac{1}{ n^{2} } - \frac{1}{2} \sum _ {n=1} ^ \infty [ \frac{1}{n} - \frac{1}{n-2} ]= \zeta (2)- \frac{1}{2} H_2= \frac{ \pi ^{2} }{6} - \frac{3}{4} $$