How to find $\int_{0}^{1}\frac{1}{x^{2}+2x+2}\mathrm dx$ with contour integration

Question

I want to evaluate the following integral $$\int_{0}^{1}\frac{1}{x^{2}+2x+2}\mathrm dx$$ by contour integration; I have a problem with the choice of the contour/ branch cuts.

Where can I find some some solved integrals like this ?

Answer 1

While I certainly wouldn't recommend residue calculus for the integral in question, I couldn't resist doing it anyway.

Let us start by a change of coordinates, mapping $[0,1]$ onto $[0,\infty)$ to get into more familiar territory for residue calculus. We can for example put $t=x/(1-x)$ to get \begin{align} \int_0^1 \frac{1}{(x+1)^2 + 1}\,dx &= \int_0^\infty \frac{1}{(\frac{t}{t+1}+1)^2 + 1}\cdot \frac{1}{(t+1)^2}\,dt \\ &= \int_0^\infty \frac{1}{5t^2+6t+2}\,dt, \\ \end{align}

and this can be evaluated by integrating $$f(z) = \frac{\log z}{5z^2+6z+2}$$ over a keyhole contour, using the natural branch of $\log z$. The function $f$ has poles at $-3/5 \pm i/5$. Omitting lots of details, you end up with

\begin{align} \int_0^\infty \frac{1}{5t^2+6t+2}\,dt &= -\operatorname{Res}\limits_{z=-3/5+i/5} \frac{\log z}{5z^2+6z+2} - \operatorname{Res}\limits_{z=-3/5-i/5} \frac{\log z}{5z^2+6z+2} \\ &= -\frac{\log(-3/5+i/5)}{2i} + \frac{\log(-3/5-i/5)}{2i} \\ &= \arctan\frac13 \end{align} which in fact is the same as $\arctan 2- \frac{\pi}{4}$ obtained by Dan (after he fixes his typo).

Answer 2

While I can help you with this particular example, if you're just looking for a reference request, I'd recommend either Lars Ahlfors's Complex Analysis or Elias Stein's Complex Analysis No. 2 (part 2 of a great 3 part series). Both contain several examples of how to integrate rational function using different types of contour integrals and try to show general methods of solution. (Although, like most types of integration, it's definitely not an exact science; there are frequently clever manipulations/argument you can do to simplify the process of integrating a particular function.)

Edit: In this particular case, it's worth noting there's a far more direct approach than contour integration: Completing the square yields

\begin{align*} \int_0^1 \frac{1}{(x + 1)^2 + 1} \, dx &= \arctan(x+1)\,\big|_{\,0}^{\,1} = \arctan(2) - \frac{\pi}{4} \end{align*}