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Evaluate:

$$\int_{-\infty}^{\infty} \frac{\log(1+x^2) dx}{1+x^2}$$

Using complex analysis, contour integration.

This function has no poles at all.

Try the contour $C$

Contoru

Obviously,

$$\oint_{C} f(z) dz = 0$$

$$\oint_{C} f(z) dz = \int_{A} f(z) dz + \int_{-R}^{R} f(x) dx$$

$$\int_{-R}^{R} f(x) dx = -\int_{A} f(z) dz$$

Along the semi circle $A$ contour-part the parametrization is:

$$z = Re^{i\theta}$$

$$\int_{A} f(z) dz = \int_{0}^{\pi} (iRe^{i\theta})\cdot \frac{\log(Re^{i\theta} + i) + \log(Re^{i\theta} - i)}{(Re^{i\theta} + i)(Re^{i\theta} - i)} d\theta$$

$$\int_{-R}^{R} f(x) dx = (-)\cdot\int_{0}^{\pi} (iRe^{i\theta})\cdot \frac{\log(Re^{i\theta} + i) + \log(Re^{i\theta} - i)}{(Re^{i\theta} + i)(Re^{i\theta} - i)} d\theta$$

$$\int_{-\infty}^{\infty} f(x) dx = (-)\cdot \lim_{R \to \infty} \int_{0}^{\pi} (iRe^{i\theta})\cdot \frac{\log(Re^{i\theta} + i) + \log(Re^{i\theta} - i)}{(Re^{i\theta} + i)(Re^{i\theta} - i)} d\theta$$

Lets say we can apply the dominated convergence theorem. We can then take the limit INSIDE the integral on the RHS. The problem becomes:

$$\int_{-\infty}^{\infty} f(x) dx = (-i)\cdot \int_{0}^{\pi} \lim_{R \to \infty} (Re^{i\theta})\cdot \frac{\log(Re^{i\theta} + i) + \log(Re^{i\theta} - i)}{(Re^{i\theta} + i)(Re^{i\theta} - i)} d\theta$$

Wolframalpha returns it as $0$ which is wrong. What is the issue? The answer is:

$$I = -\pi\log(\sqrt{2})$$

Please tell me what is wrong with this, do not suggest something else until the end. Thank you!

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Your integrand has branch points within the contour you are integrating, so Cauchy's theorem does not apply there. You have to exclude the branch points of your integrand. You may do so by deforming the contour along the imaginary axis to encircle the branch point.

See, for example, this answer.

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