I want to calculate $$I=\int_{0}^{\infty}\dfrac{\ln x}{x^2+\alpha^2}\,\mathrm{d}x.$$

my first attempt:

$$I(t)=\displaystyle\int_{0}^{\infty}\dfrac{\ln {tx}}{x^2+\alpha^2}\,\mathrm{d}x$$

$$I'(t)=\dfrac{1}{t}\displaystyle\int_{0}^{\infty}\dfrac{1}{x^2+\alpha^2}\,\mathrm{d}x=\dfrac{\pi}{2\alpha t}\implies I(t)=\dfrac{\pi\ln t }{2\alpha}\ + C$$

But I'm unable to calculate $C$ because $I(0)=\infty$

so,please give me hint to calculate $C$

or hint me to stick parameter $t$ in right place

my second attempt (but without differentiation under integral sign):

I substituted $x=\alpha \tan t$

$I=\displaystyle \int _{0}^{\dfrac{\pi}{2}} \dfrac{\ln\alpha \tan t}{\alpha^2 \sec^2 t}\alpha \sec^2 t dt$

$I=\displaystyle\dfrac{1}{\alpha} \int _{0}^{\dfrac{\pi}{2}}{\ln\alpha \tan t} dt$

$I=\displaystyle\dfrac{1}{\alpha} \int _{0}^{\dfrac{\pi}{2}}{\ln \alpha} dt +\displaystyle\dfrac{1}{\alpha} \int _{0}^{\dfrac{\pi}{2}}\ln\tan t dt $

$I=\dfrac{1}{\alpha}\dfrac{\pi}{2}\ln \alpha+0$


but this question was given in my coaching sheet under the topic DUIS (Differentiation under integral sign) so, i want to do it by that technique only, any help will be appreciated, thank you ..

  • 2
    $\begingroup$ Hint compare $I(1)$ and $I(2)$, say, to obtain $C$. $\endgroup$ – user431008 Feb 11 '18 at 19:02
  • 1
    $\begingroup$ Also $I(0^+) = -\infty$ not $\infty$ $\endgroup$ – Von Neumann Feb 11 '18 at 19:17
  • $\begingroup$ Sorry, my post was true only for $\alpha=1$ $\endgroup$ – Atmos Feb 11 '18 at 19:25

Let $F(a)$ be given by the integral

$$\begin{align} F(a)&=\int_0^\infty \frac{x^a}{x^2+\alpha^2}\,dx\\\\ &=\alpha^{a-1} \int_0^\infty \frac{x^a}{x^2+1}\,dx\tag1 \end{align}$$

Next, enforce the substitution $x\mapsto e^x$ to obtain

$$\begin{align} \int_0^\infty \frac{x^a}{1+x^2}\,dx&=\int_{-\infty}^\infty \frac{e^{(a+1)x}}{1+e^{2x}}\,dx\\\\ &=\int_{-\infty}^0\frac{e^{(a+1)x}}{1+e^{2x}}\,dx+\int_{0}^\infty\frac{e^{(a-1)x}}{1+e^{-2x}}\,dx\\\\ &=\sum_{n=0}^\infty (-1)^n\left(\int_{-\infty}^0 e^{(2n+1+a)x}\,dx+\int_{0}^\infty e^{-(2n+1-a)x}\,dx\right)\\\\ &=\sum_{n=0}^\infty (-1)^n \left(\frac{1}{2n+1+a}+\frac{1}{2n+1-a}\right)\\\\ &=2\sum_{n=0}^\infty (-1)^n\left(\frac{2n+1}{(2n+1)^2-a^2}\right) \\\\ &=\frac{\pi}{2}\sec\left(\frac{\pi a}{2}\right)\tag 2 \end{align}$$

See THIS ANSWER along with other answers on that page and THIS ANSWER for an alternative approaches to evaluating $\displaystyle \int_0^\infty \frac{x^a}{x^2+1}\,dx$.

Substituting $(2)$ into $(1)$ reveals

$$F(a)=\alpha^{a-1}\frac{\pi}{2}\sec\left(\frac{\pi a}{2}\right)\tag 3$$

Differentiating $(3)$ and setting $a=0$ yields

$$F'(0)=\int_0^\infty \frac{\log(x)}{x^2+\alpha^2}\,dx=\frac{\pi \log(\alpha)}{2\alpha}$$

  • $\begingroup$ @ Mark Viola...:)fantastic answer Sir ,.you brilliantly used G.P.(geometric progression) and taylor series for sec x but not the "differentiation under integral sign"......i mean is it approachable that way also(using differentiation under integral sign) ? $\endgroup$ – Faraday Pathak Feb 11 '18 at 19:56
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    $\begingroup$ @veereshpandey Thank you. But, I did use differentiation under the integral in the last step of the development. Differentiation under the integral was not used to arrive at $F(a)$. $\endgroup$ – Mark Viola Feb 11 '18 at 20:52

It is enough to exploit symmetry. Assuming $\alpha>0$ we have $$ \int_{0}^{+\infty}\frac{\log x}{x^2+\alpha^2}\,dx \stackrel{x\mapsto\alpha z}{=} \frac{1}{\alpha}\int_{0}^{+\infty}\frac{\log\alpha+\log z}{z^2+1}\,dz=\color{red}{\frac{\pi\log\alpha}{2\alpha}}+\frac{1}{\alpha}\int_{0}^{+\infty}\frac{\log z}{z^2+1}\,dz$$ and the last integral equals zero by mapping $z$ into $\frac{1}{w}$.

  • 1
    $\begingroup$ Jack, of course the development is efficient. But it doesn't rely on Feynman's trick, which was the way forward the OP had sought. $\endgroup$ – Mark Viola Feb 12 '18 at 4:04

Here is an alternative complex method that still uses differentiation under the integral sign. Denote $$I(b)=\int_0^\infty \frac{x^b}{x^2+\alpha^2}\,dx=\frac{1}{2}\int_{-\infty}^\infty \frac{x^b}{x^2+\alpha^2}\,dx$$ Assuming that $b=2\beta$, for $\beta\in\mathbb{R}$. Now, $b$ is restricted to certain values to ensure convergence, and we will choose $b\in[0,1)$. We are then trying to compute $I'(0)$. We will evaluate $I(b)$ via the Residue Theorem, differentiate the resulting function, and let $b\to 0$.

Using a semi-circle centered at the origin, its contour integral is given by the upper semi-circle, parametrized by $z=Re^{i\theta},\,\,\theta \in[0, \pi]$ added to the real contribution, parametrized by $z=x,\,\,x\in[-R,R]$. $$ \int_{-R}^R \frac{x^b}{x^2+\alpha^2}\,dx\, +\, \int_0^{\pi}\frac{R^be^{ib\theta}}{R^2e^{2i\theta}+\alpha^2}Rie^{i\theta}\,d\theta$$ Because of our choice of $b$, the second integral vanishes as $R\to\infty$. The first integral is equal to $2I(b)$ in the same limit. Using the Residue Theorem, $$\int_{-\infty}^{\infty} \frac{x^b}{x^2+\alpha^2}\,dx=2\pi i\,\text{Res}\left(z=\alpha i\right) \frac{z^b}{z^2+\alpha^2}\ $$ As the singularity of our function inside our semi circle occurrs at $z=\alpha i$. The Residue is equal to $$\text{Res}=\lim_{z\to\alpha i} \,\,\,(z-\alpha i)\frac{z^b}{(z-\alpha i)(z+\alpha i)}=\frac{\alpha^b i^b}{2\alpha i}$$ Thus, $$2I(b)=2\pi i \frac{\alpha^b i^b}{2\alpha i}$$ $$I(b)=\frac{\pi}{2\alpha}\alpha^b i^b$$ Differentiating both sides with repect to $b$, $$I'(b)=\frac{\pi}{2\alpha} \left(\alpha^b i^b \log \alpha +\alpha^bi^b\frac{i\pi}{2}\right)$$ Letting $b\to0$, and taking the real part of our expression, we have $$\int_0^{\infty}\frac{\log x}{x^2+\alpha^2}\,dx=\frac{\pi}{2\alpha}\log \alpha$$


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