# Thread: Prove the improper integral below

1. ## Prove the improper integral below

Prove that

$\displaystyle \int_0^{\infty}\frac{\ln(x)}{1+x^s}dx=-\left(\frac{\pi}{s}\right)^2\cdot{\cot\left(\frac{ \pi}{s}\right)\cdot{\text{cosec}\left(\frac{\pi}{s }\right)}}$

$\displaystyle s > 1$

2. Originally Posted by mathwizard
Prove that

$\displaystyle \int_0^{\infty}\frac{\ln(x)}{1+x^s}dx=-\left(\frac{\pi}{s}\right)^2\cdot{\cot\left(\frac{ \pi}{s}\right)\cdot{\text{cosec}\left(\frac{\pi}{s }\right)}}$

$\displaystyle s > 1$
haha, that's a question for PaulRS. it's his signature

3. $\displaystyle I = \int_0^\infty {\tfrac{{\ln \left( x \right)}} {{1 + x^a }}dx} = \int_0^\infty {\ln \left( x \right)\left( {\int_0^\infty {e^{ - y \cdot \left( {1 + x^a } \right)} dy} } \right)dx} $$\displaystyle = \int_0^\infty {e^{ - y} \cdot \left( {\int_0^\infty {\ln \left( x \right)e^{ - y \cdot x^a } dx} } \right)dy} Note that: \displaystyle \Gamma \left( s \right) = \int_0^\infty {x^{s - 1} \cdot e^{ - x} dx} \Rightarrow \Gamma '\left( s \right) = \int_0^\infty {e^{ - x} \cdot \tfrac{{\partial x^{s - 1} }} {{\partial s}}dx} = \int_0^\infty {\ln \left( x \right) \cdot x^{s - 1} \cdot e^{ - x} dx} Now: \displaystyle \sqrt[a]{y} \cdot \int_0^\infty {\ln \left( {\sqrt[a]{y} \cdot x} \right)e^{ - \left( {\sqrt[a]{y} \cdot x} \right)^a } dx} = \int_0^\infty {\ln \left( x \right)e^{ - x^a } dx} and: \displaystyle \int_0^\infty {\ln \left( x \right)e^{ - x^a } dx} \underbrace = _{x^a = u}\tfrac{1} {{a^2 }} \cdot \int_0^\infty {\ln \left( u \right) \cdot u^{\tfrac{1} {a} - 1} \cdot e^{ - u} d} u = \tfrac{1} {{a{}^2}} \cdot \Gamma '\left( {\tfrac{1} {a}} \right) Thus we have, by properties of the loagrithm: \displaystyle \sqrt[a]{y} \cdot \int_0^\infty {\ln \left( {\sqrt[a]{y} \cdot x} \right)e^{ - \left( {\sqrt[a]{y} \cdot x} \right)^a } dx} =$$\displaystyle \tfrac{{\sqrt[a]{y} \cdot \ln \left( y \right)}} {a} \cdot \int_0^\infty {e^{ - \left( {\sqrt[a]{y} \cdot x} \right)^a } dx} + \sqrt[a]{y} \cdot \int_0^\infty {\ln \left( x \right)e^{ - \left( {\sqrt[a]{y} \cdot x} \right)^a } dx}$

So that: $\displaystyle \sqrt[a]{y} \cdot \int_0^\infty {\ln \left( {\sqrt[a]{y} \cdot x} \right)e^{ - \left( {\sqrt[a]{y} \cdot x} \right)^a } dx} =$$\displaystyle \tfrac{{\ln \left( y \right)}} {{a^2 }} \cdot \Gamma \left( {\tfrac{1} {a}} \right) + \sqrt[a]{y} \cdot \int_0^\infty {\ln \left( x \right)e^{ - \left( {\sqrt[a]{y} \cdot x} \right)^a } dx}$

and: $\displaystyle \int_0^\infty {\ln \left( x \right)e^{ - \left( {\sqrt[a]{y} \cdot x} \right)^a } dx} = \frac{{\tfrac{1} {{a{}^2}} \cdot \Gamma '\left( {\tfrac{1} {a}} \right) - \tfrac{{\ln \left( y \right)}} {{a^2 }} \cdot \Gamma \left( {\tfrac{1} {a}} \right)}} {{\sqrt[a]{y}}}$

Integrating: $\displaystyle I = \int_0^\infty {e^{ - y} \cdot \left( {\frac{{\tfrac{1} {{a{}^2}} \cdot \Gamma '\left( {\tfrac{1} {a}} \right) - \tfrac{{\ln \left( y \right)}} {{a^2 }} \cdot \Gamma \left( {\tfrac{1} {a}} \right)}} {{\sqrt[a]{y}}}} \right)dy} = \frac{{\Gamma '\left( {\tfrac{1} {a}} \right)\Gamma \left( {1 - \tfrac{1} {a}} \right) - \Gamma \left( {\tfrac{1} {a}} \right)\Gamma '\left( {1 - \tfrac{1} {a}} \right)}} {{a^2 }}$

By Euler's Reflection Formula: $\displaystyle \Gamma \left( x \right) \cdot \Gamma \left( {1 - x} \right) = \frac{\pi } {{\sin \left( {\pi \cdot x} \right)}} \Rightarrow \Gamma '\left( x \right) \cdot \Gamma \left( {1 - x} \right) - \Gamma \left( x \right) \cdot \Gamma '\left( {1 - x} \right) = - \frac{{\pi ^2 \cdot \cos \left( {\pi \cdot x} \right)}} {{\sin ^2 \left( {\pi \cdot x} \right)}}$

Thus: $\displaystyle {I = - \left( {\tfrac{\pi } {a}} \right)^2 \cdot \frac{{\cos \left( {\tfrac{\pi } {a}} \right)}} {{\sin ^2 \left( {\tfrac{\pi } {a}} \right)}} = - \left( {\tfrac{\pi } {a}} \right)^2 \cdot \cot \left( {\tfrac{\pi } {a}} \right) \cdot \text{cosec} \left( {\tfrac{\pi } {a}} \right)}$