Originally Posted by

**arbolis** I'm having a textbook at hand (Complex Analysis by Sakarchi, Stein) and I don't understand something.

At page 78, he wants to prove that $\displaystyle \int _{- \infty}^{\infty} \frac{dx}{1+x^2}= \pi$.

He says that $\displaystyle f(z)=\frac{1}{z^2+1}$ has 2 simple poles, one at $\displaystyle -i$ and the other at $\displaystyle i$. Until now everything's fine.

Now in order to calculate the integral, he chose a contour $\displaystyle \gamma _R$ where this curve is the half circle with radius R, centered at the origin in the upper half plane.

He then says that the residue of $\displaystyle f$ at 1 is $\displaystyle \frac{1}{2i}$, I still follow him.

Now he says that $\displaystyle \int _{\gamma _R} f(z)dz=\frac{2 \pi i}{2i}=\pi$. That's fine since the pole at 1 is enclosed into the curve $\displaystyle \gamma _R$.

**Now this is where I don't understand. **

He goes on to say "let C be the large half circle of radius R. We see that $\displaystyle \left | \int _ C f(z)dz \right | \leq \pi R \frac{B}{R^2} \leq \frac{M}{R}$. The integral goes to $\displaystyle 0$ as $\displaystyle R \to \infty$. Therefore in the limit we find $\displaystyle \int _{- \infty}^{\infty} \frac{dx}{1+x^2}= \pi $".

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1)He never defined B and M (at least in the previous pages). I guess that they are arbitrary.

2)I don't see the necessity of finding a upper bound of the integral. (Why did he did so?)

3)He never found a curve enclosing the other pole! So how can the integral whose limits are $\displaystyle -\infty$ and $\displaystyle \infty$ be calculated via the residue theorem for a single pole while the function has 2 poles? Is it because Sakarchi's curve encloses all real numbers when R tends to $\displaystyle + \infty$?

If so then I could simply chose a curve enclosing all real numbers when I take the limit, but not enclosing any poles, and the contour integral would be worth $\displaystyle 0$ according to Cauchy's integral theorem. Which makes no sense since the integral is worth $\displaystyle \pi$.

Thanks for reading and helping.