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PostPosted: Wed Jun 07, 2017 11:11 am 
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For $0<\alpha<\beta$, evaluate
\[\displaystyle\int_{0}^{+\infty}\frac{\log{x}}{(x+\alpha)(x+\beta)}\,dx\,.\]

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PostPosted: Sat Jun 10, 2017 9:12 pm 
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Let \(\displaystyle{F(a,b)=\int_{0}^{\infty}\dfrac{\ln\,x}{(x+a)\,(x+b)}\,\mathrm{d}x\,\,,0<a<b}\)

Using the substituton, we get

\(\displaystyle{F(a,b)=-\int_{\infty}^{0}\dfrac{\ln(1/t)\,t^2}{(1+a\,t)\,(1+b\,t)}\,\dfrac{1}{t^2}\,\mathrm{d}t=\int_{0}^{\infty}\dfrac{-\ln\,x}{(a\,x+1)\,(b\,x+1)}\,\mathrm{d}x\,\,(I)}\)

Futhermore,

\(\displaystyle{\begin{aligned}F(a,b)&=\int_{0}^{\infty}\dfrac{\ln\,(a\,b)+\ln\,(x/(a\,b))}{a\,b\,\left(x/a+1\right)\,\left(x/b+1\right)}\,\mathrm{d}x\\&\stackrel{y=x/a\,b}{=}\int_{0}^{\infty}\dfrac{\ln\,(a\,b)+\ln\,x}{(a\,x+1)\,(b\,x+1)}\,\mathrm{d}x\,\,(II) \end{aligned}}\)

According to the relations \(\displaystyle{(I)\,,(II)}\) we have that

\(\displaystyle{\begin{aligned}2\,F(a,b)&=\int_{0}^{\infty}\dfrac{\ln\,(a\,b)}{(a\,x+1)\,(b\,x+1)}\,\mathrm{d}x\\&=\ln\,(a\,b)\,\int_{0}^{\infty}\left(\dfrac{b}{b-a}\,\dfrac{1}{b\,x+1}-\dfrac{a}{b-a}\,\dfrac{1}{a\,x+1}\right)\,\mathrm{d}x\\&=\left[\dfrac{\ln\,(a\,b)}{b-a}\,\ln\,\left(\dfrac{b\,x+1}{a\,x+1}\right)\right]_{0}^{\infty}\\&=\dfrac{1}{b-a}\,\ln\,(a\,b)\,\ln\,(b/a) \end{aligned}}\)

so

\(\displaystyle{F(a,b)=\dfrac{1}{2\,(b-a)}\,\left(\ln^2\,b-\ln^2\,a\right)}\)


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PostPosted: Sat Jun 10, 2017 9:22 pm 
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Another solution using complex analysis:

We consider the complex function $f(z)=\frac{{\rm{Log}}^2{z}}{(z-\alpha)(z-\beta)}\,,\quad z\in\mathbb{C}\setminus\{x+0i\;|\;x\leqslant0\}$, which is meromorphic at \({\mathbb{C}}\setminus\{x+0i\;|\;x\leqslant0\}\) with simple poles $z=\alpha$, $z=\beta$ and the simple, closed curve \(\gamma\), which is the summand of the anticlockwise oriented arc \(C_r:=\big\{r\,{\rm{e}}^{it}\in{\mathbb{C}}\;\big|\; t\in(-\pi+\varepsilon,\pi-\varepsilon]\big\}\,, \; r\to+\infty\,,\;\varepsilon\to0^+\,, \) of the segment \(\ell:=\big\{(r+\delta-t)\,{\rm{e}}^{i\pi}\in{\mathbb{C}}\;\big|\; t\in(\delta,r)\big\}\), of the clockwise oriented arc \(C_\delta:=\big\{\delta\,{\rm{e}}^{-it}\in{\mathbb{C}}\;\big|\; t\in(-\pi+\varepsilon,\pi-\varepsilon]\big\}\,, \;\delta\to0^{+}\,,\;\varepsilon\to0^+\,,\) and of the segment \(-\ell:=\big\{t\,{\rm{e}}^{-i\pi}\in{\mathbb{C}}\;\big|\; t\in(\delta,r)\big\}\).

Attachment:
contour.png



We have that \begin{align*}
{\rm{Res}\,}\big(f(z), z=\alpha\big)&=\frac{1}{0!}\,\mathop{\lim}\limits_{z\to\alpha}\bigg(( z-\alpha)\,\dfrac{{\rm{Log}\,}^2{z}}{(z-\alpha)(z-\beta)}\bigg)\\
&=\mathop{\lim}\limits_{z\to\alpha}\dfrac{{\rm{Log}\,}^2{z}}{z-\beta}\\
&= \dfrac{\log^2{\alpha}}{\alpha-\beta}\,,\\
{\rm{Res}\,}\big(f(z), z=\beta\big)&=\frac{1}{0!}\,\mathop{\lim}\limits_{z\to\beta}\bigg(( z-\beta)\,\dfrac{{\rm{Log}\,}^2{z}}{(z-\alpha)(z-\beta)}\bigg)\\
&=\mathop{\lim}\limits_{z\to\beta}\dfrac{{\rm{Log}\,}^2{z}}{z-\alpha}\\
&=\dfrac{\log^2{\beta}}{\beta-\alpha}
\end{align*} and \begin{align*}
\displaystyle\oint_{\gamma}{f(z)\,dz}&=2\pi i\;\cancelto{1}{{\rm{I}\,}\big(\gamma, z=\alpha\big)}\,{\rm{Res}\,}\big(f(z), z=\alpha\big)+2\pi i\;\cancelto{1}{{\rm{I}\,}\big(\gamma, z=\beta\big)}\,{\rm{Res}\,}\big(f(z), z=\beta\big)\\
&=2\pi i\,\dfrac{\log^2{\alpha}}{\alpha-\beta}+2\pi i\,\dfrac{\log^2{\beta}}{\beta-\alpha}\\
&=2\pi i\,\dfrac{\log^2\beta-\log^2\alpha}{\beta-\alpha}&\Longrightarrow\\
\oint_{C_{r}}{f(z)\,dz}&+\oint_{\ell}{f(z)\,dz}+\oint_{C_{\delta}}{f(z)\,dz}+\oint_{-\ell}{f(z)\,dz}=2\pi i\,\dfrac{\log^2\beta-\log^2\alpha}{\beta-\alpha}\qquad(1)\,.
\end{align*}
Also \begin{align*}
\displaystyle\oint_{\ell}{f(z)\,dz}&=\int_{\delta}^{r}{\dfrac{{\rm{Log}\,}^2\big((r+\delta-t)\,{\rm{e}}^{i\pi}\big)\,\big((r+\delta-t)\,{\rm{e}}^{i\pi}\big)'}{\big((r+\delta-t)\,{\rm{e}}^{i\pi}-\alpha\big)\big((r+\delta-t)\,{\rm{e}}^{i\pi}-\beta\big)}\,dt}\\
&=\int_{\delta}^{r}{\dfrac{\big(\log|-r-\delta+t|+i\,\pi\big)^2}{(-r-\delta+t-\alpha)(-r-\delta+t-\beta)}\,dt}\\
&\mathop{=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{s\,=\,r+\delta-t}\\
{ds\,=\,-dt}\\
\end{subarray}}\,\int_{\delta}^{r}{\dfrac{\big(\log{s}+i\,\pi\big)^2}{(-s-\alpha)(-s-\beta)}\,ds}\\
&=\int_{\delta}^{r}{\dfrac{\big(\log{t}+i\,\pi\big)^2}{(t+\alpha)(t+\beta)}\,dt}\,,\\
\displaystyle\oint_{-\ell}{f(z)\,dz}&=\int_{\delta}^{r}{\dfrac{{\rm{Log}\,}^2\big(t\,{\rm{e}}^{-i\pi}\big)\,\big(t\,{\rm{e}}^{-i\pi}\big)'}{\big(t\,{\rm{e}}^{-i\pi}-\alpha\big)\big(t\,{\rm{e}}^{-i\pi}-\beta\big)}\,dt}\\
&=-\int_{\delta}^{r}{\dfrac{\big(\log|{-t}|-i\,\pi\big)^2}{(-t-\alpha)(-t-\beta)}\,dt}\\
&=-\int_{\delta}^{r}{\dfrac{\big(\log{t}-i\,\pi\big)^2}{(t+\alpha)(t+\beta)}\,dt}\hspace{3.0cm}\Rightarrow\\
\int_{\ell}{f(z)\,dz}+\int_{-\ell}{f(z)\,dz}&=\int_{\delta}^{r}{\dfrac{\big(\log{t}+i\,\pi\big)^2}{(t+\alpha)(t+\beta)}\,dt}-\int_{\delta}^{r}{\dfrac{\big(\log{t}-i\,\pi\big)^2}{(t+\alpha)(t+\beta)}\,dt}\\
&=\int_{\delta}^{r}{\dfrac{4\pi\,i\,\log{t}}{(t+\alpha)(t+\beta)}\,dt}\,.
\end{align*}
By Estimation lemma we have that
\[\displaystyle\mathop{\lim}\limits_{r\to+\infty}\oint_{C_{r}}{f(z)\,dz}=\mathop{\lim}\limits_{\delta\to0^{+}}\oint_{C_{\delta}}{f(z)\,dz}=0\,.\]
So equation $(1)$ becames \begin{align*}
\mathop{\lim}\limits_{\begin{subarray}{c}
{r\to+\infty}\\
{\delta\to0^+}
\end{subarray}}\int_{\delta}^{r}{\dfrac{4\pi\,i\,\log{t}}{(t+\alpha)(t+\beta)}\,dt}+\cancelto{0}{\mathop{\lim}\limits_{\begin{subarray}{c}
{r\to+\infty}\\
{\delta\to0^+}
\end{subarray}}\int_{C_{r}}{f(z)\,dz}}+\cancelto{0}{\mathop{\lim}\limits_{\begin{subarray}{c}
{r\to+\infty}\\
{\delta\to0^+}
\end{subarray}}\int_{C_{\delta}}{f(z)\,dz}}&=2\pi i\,\dfrac{\log^2\beta-\log^2\alpha}{\beta-\alpha} &\Rightarrow\\
\mathop{\lim}\limits_{\begin{subarray}{c}
{r\to+\infty}\\
{\delta\to0^+}
\end{subarray}}\int_{\delta}^{r}{\dfrac{\log{t}}{(t+\alpha)(t+\beta)}\,dt}&=\dfrac{\log^2\beta-\log^2\alpha}{2\,(\beta-\alpha)} &\Rightarrow\\
\displaystyle\int_{0}^{+\infty}{\dfrac{\log{t}}{(t+\alpha)(t+\beta)}\,dt}=\dfrac{\log^2\beta-\log^2\alpha}{2\,(\beta-\alpha)}\,.
\end{align*}

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