Evaluate the following integrals: ^ _0 1+x^2 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\limits^{\infty}_0\frac{\log\text{x}}{1+\text{x}^2}\text{ dx}$

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Answer

We have,
$\text{I}=\int\limits^{\infty}_0\frac{\log\text{x}}{1+\text{x}^2}\text{ dx}$
Putting $\text{x}=\tan\theta$
$\text{dx}=\sec^2\theta\text{ d}\theta$
When $\text{x}\rightarrow0;\theta\rightarrow0$
And $\text{x}\rightarrow\infty;\theta\rightarrow\frac{\pi}{2}$
Now, integral becomes,
$\therefore\ \text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\log(\tan\theta)}{1+\tan^2\theta}\sec^2\theta\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\log(\tan\theta)\text{d}\theta\ ...(\text{i})$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\Big[\tan\Big(\frac{\pi}{2}-\theta\Big)\Big]\text{d}\theta$ $\Bigg[\because\ \int\limits^{\text{a}}_{0}\text{f(x)}\text{dx}=\int\limits^{\text{a}}_{\text{0}}\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\log(\cot\theta)\text{d}\theta\ ....(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\log(\tan\theta)\text{d}\theta+\int\limits^{\frac{\pi}{2}}_0\log(\cot\theta)\text{d}\theta$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\big[\log(\tan\theta)+\log(\cot\theta)\big]\text{d}\theta$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\big[\log(\tan\theta\times\cot\theta)\big]\text{d}\theta$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\big(\log1\big)\text{d}\theta$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0(0)\text{d}\theta$
$\Rightarrow2\text{I}=0$
$\Rightarrow\text{I}=0$
$\therefore\ \int\limits^{\infty}_0\frac{\log\text{x}}{1+\text{x}^2}\text{ dx}=0$

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