\(\int_{0}^{+\infty}\frac{\log^2{x}}{(x+\alpha)(x+\beta)}\,dx\)
- Grigorios Kostakos
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- Location: Ioannina, Greece
\(\int_{0}^{+\infty}\frac{\log^2{x}}{(x+\alpha)(x+\beta)}\,dx\)
As a continuation of this:
For $0<\alpha<\beta$, evaluate
\[\displaystyle\int_{0}^{+\infty}\frac{\log^2{x}}{(x+\alpha)(x+\beta)}\,dx\,.\]
For $0<\alpha<\beta$, evaluate
\[\displaystyle\int_{0}^{+\infty}\frac{\log^2{x}}{(x+\alpha)(x+\beta)}\,dx\,.\]
Grigorios Kostakos
- Grigorios Kostakos
- Founder
- Posts: 461
- Joined: Mon Nov 09, 2015 1:36 am
- Location: Ioannina, Greece
Re: \(\int_{0}^{+\infty}\frac{\log^2{x}}{(x+\alpha)(x+\beta)}\,dx\)
For the sake of discussion I give the next step:
Considering the complex function $$f(z)=\dfrac{{\rm{Log}}^3{z}+\pi^2{\rm{Log}}{z}}{(z-\alpha)(z-\beta)}\,,\quad z\in\mathbb{C}\setminus\{x+0i\;|\;x\leqslant0\},$$ by the same procedure as in \(\int_{0}^{+\infty}\frac{\log{x}}{(x+\alpha)(x+\beta)}\,dx\), we will get
\[\displaystyle\int_{0}^{+\infty}\frac{\log^2{x}}{(x+\alpha)(x+\beta)}\,dx=\dfrac{\log^3\beta+\pi^2\log\beta-\log^3\alpha-\pi^2\log\alpha}{3\,(\beta-\alpha)}\,.\]
Can you find appropriate complex functions for higher powers of \(\log{x}\) i.e. \(\log^3{x}\,,\; \log^4{x}\,,\) ...?
Considering the complex function $$f(z)=\dfrac{{\rm{Log}}^3{z}+\pi^2{\rm{Log}}{z}}{(z-\alpha)(z-\beta)}\,,\quad z\in\mathbb{C}\setminus\{x+0i\;|\;x\leqslant0\},$$ by the same procedure as in \(\int_{0}^{+\infty}\frac{\log{x}}{(x+\alpha)(x+\beta)}\,dx\), we will get
\[\displaystyle\int_{0}^{+\infty}\frac{\log^2{x}}{(x+\alpha)(x+\beta)}\,dx=\dfrac{\log^3\beta+\pi^2\log\beta-\log^3\alpha-\pi^2\log\alpha}{3\,(\beta-\alpha)}\,.\]
Can you find appropriate complex functions for higher powers of \(\log{x}\) i.e. \(\log^3{x}\,,\; \log^4{x}\,,\) ...?
Grigorios Kostakos
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