Integrating $\int_{0}^{1}\frac{dz}{1+z^2}$

Question

I am trying to calculate

$$\int_{0}^{1}\frac{dz}{1+z^2}$$

But I am unsure how to interpret the integral, does this mean integrating the function along any path starting at $0$ and ending at $1$?

Answer 1

As others have remarked the complex integral $$\int_0^1{dz\over 1+z^2}$$ is not well defined unless you specify the intended path of integration. If this path is just the segment $$\sigma:\quad t\mapsto t\in{\mathbb C}\qquad(0\leq t\leq1)$$ then of course $$\int_\sigma{dz\over 1+z^2}=\int_0^1{dt\over 1+t^2}={\pi\over 4}\ .$$ If, however, $\gamma$ is some arbitrary path beginning at $0$ and ending at $1\in{\mathbb C}$, then we can argue as follows: The path (chain) $\delta:=\gamma-\sigma$ is closed. If $\delta$ avoids the points $\pm i$, which we shall assume here, then by the general residue theorem we have $$\int_\delta{dz\over 1+z^2}=2\pi i\bigl(n(\delta, i)\>{\rm res}(f,i) - n(\delta, -i)\>{\rm res}(f,-i)\bigr)\ .$$

The function $$f(z):={1\over 1+z^2}={1\over (z-i)(z+i)}$$ has residues $\pm{1\over2i}$ at $\pm i$. It follows that $$\int_\delta{dz\over 1+z^2}=\pi\bigl(n(\delta, i) - n(\delta, -i)\bigr)\ ,$$ so that we finally obtain $$\int_\gamma{dz\over 1+z^2}=\int_\sigma{dz\over 1+z^2}+\int_\delta{dz\over 1+z^2}={\pi\over 4}+ \pi\bigl(n(\delta, i) - n(\delta, -i)\bigr)\ .$$

Answer 2

it means $$\int_0^1 \frac{1}{1+z^2} dz$$ and we can remember that $$\frac{d}{dx} \arctan x=\frac{1}{1+x^2}$$

Answer 3

This notation should be avoided. The most reasonable interpretation is the one you give, but in this particular case, the value of the integral is not independent of the chosen path.

Taking a path that winds around $z=i$ or $z=-i$ may result in a different value.