Riemann’s zeta function is defined as \(\zeta(z) = \sum_{n=1}^\infty n^{-z}\).
For which values of \(z\) does this converge?
Show that the zeta function admits the integral representation \[ \zeta(z) = \frac{1}{\Gamma(z)} \int_0^\infty \frac{t^{z-1}}{e^t - 1}\,\mathrm{d}t \] hint: the relation \(\sum_{m=1}^\infty e^{-mt} = \frac{e^{-t}}{1 - e^{-t}}\) might prove useful.
Now we take the contour of figure 5. Since we want to allow non integer values of \(z\), there is a branch cut along the positive real axis. What is \(\frac{1}{\Gamma(z)} \int_0^\infty \frac{t^{z-1}}{e^t - 1}\,\mathrm{d}t\) along this contour?
For \(z < 0\) we can deform the contour by sending the radius of the circle \(D\) to infinity; the price to pay is that, to compute \(I\), we have to evaluate an infinite number of poles, but this can be done. By comparing this result to what you did in the previous step, you should find that \[ \zeta(z) = \zeta(1-z)\, \frac{e^{3\pi i z/2} - e^{\pi i z/2}}{e^{2\pi i z} - 1}\, \frac{(2\pi)^z}{\Gamma(z)}. \]
Using the formula you just found, show that \(\zeta(-1) = -\frac{1}{12}\). Notice that this doesn’t mean that \(1 + 2 + 3 + 4 + \ldots = -\frac{1}{12}\), since that \(\zeta(z) = \sum_{n=1}^\infty n^{-z}\) is valid only for \(z > 1\). On a side note, this is the reason why string theory (without supersymmetry) needs 26 spacetime dimensions.
Figure 5 of the source (contour for problem 32) shows, with the cut line along the positive real axis, a contour that comes in from \(+\infty\) along a line \(A\) just above the cut, goes counterclockwise around the origin along a small circle \(D\) of radius \(\varepsilon\), and returns to \(+\infty\) along a line \(B\) just below the cut.
The source uses \(I\) in the fourth part without defining it; it presumably denotes the contour integral of the third part.