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Suppose we are given:

$$\text{Evaluate} \int_{0}^{1} \frac{1}{1+x^2} \text{dx}$$

This is quite easy because you will notice that:

$$\int_{0}^{1} \frac{1}{1+x^2} \text{dx} = \arctan(1) - \arctan(0) = \frac{\pi}{4} $$

But can this be done using complex analysis, contour integration?

Also, is this possible (using contour integration?)

$$\int_{0}^{1} \frac{\ln(1+x)}{1+x^2} \text{dx}$$

Thanks!

Amad27
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1 Answers1

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But can this be done using complex analysis, contour integration?

Of course! As it has already been suggested in the comments, use the symmetry of the integrand to rewrite the integral as $~\displaystyle\frac12\int_{-1}^1\frac{dx}{1+x^2},~$ then choose $x=e^{it}$, where $t\in(-\pi,0)$, and notice that the denominator has roots in $x=\pm i\iff t=\pm\dfrac\pi2.~$ I believe you can take it from here. :-$)$

Lucian
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  • If you use that contour, since the root lies on the contour, the integral will not converge. – Doug M Aug 30 '19 at 21:44
  • @DougM: And what do we usually do when such a situation arises ? (As one can safely infer from the answer's last sentence, the finishing touches are left to the reader, rather than being spoon-fed). – Lucian Aug 30 '19 at 22:57