I've been working on this integral for quite a while and I think I've been able to progress but now I'm stuck. So I have to prove that

$$\int_C f(z)\ dz =\int_C\frac{2z}{(1+z^2)\log(2+z^2)}dz =\pi i.$$

Where $C$ is a contour from $[1,p]$ to $[p,-1]$ and $p=\frac{6i}5$.

My attempt :

  1. So I have noticed that $f(-z)=-f(z)$ and so $f$ is odd and the contour can be completed (triangle) to be closed since $\int_Df(z)dz=0$ where $D$ is a line segment from $-1$ to $1$.
  2. Clearly the only singularity in the new contour (the triangle) is $i$.

So this is where I get stuck. I've tried to use Cauchy's integral formula but the $\log(2+z^2)$ gives me hard time.

I would appreciate a hint. (Residue theorem is not allowed, almost everything else is).

  • $\begingroup$ Is $\displaystyle{\large C}$ a contour or a straight line from $\displaystyle{\large\left[1,p\right]}$ to $\displaystyle{\large\left[p,-1\right]}$ ?. $\endgroup$ – Felix Marin Aug 2 '14 at 17:40

Suggestion: Substitute $w = 1+z^2$ to simplify the situation. The image of $C$ under that substitution becomes a closed curve $\gamma$ in the half-plane $H = \{ w : \operatorname{Re} w > -1\}$ winding once around $0$, and the integral

$$\int_\gamma \frac{dw}{w\log (1+w)}.$$

Now we want to transform it into a way so that we can use the Cauchy integral formula (which is a special case of the residue theorem, but one that seems to be allowed). For that, consider the function

$$f(w) = \begin{cases}\qquad 1 &, w = 0\\ \dfrac{w}{\log (1+w)} &, w\neq 0 \end{cases}$$

which is holomorphic on $H$. By the integral formula,

$$\int_\gamma \frac{dw}{w\log (1+w)} = \int_\gamma \frac{f(w)}{w^2}\,dw = 2\pi i f'(0).$$

Of course you can play the same game without the substitution after closing $C$ with

$$F(z) = \begin{cases} \dfrac{2z(z-i)}{(z+i)\log (2+z^2)} &, z \neq i\\ \qquad \dfrac{1}{2i} &, z = i,\end{cases}$$

but it's easier to compute with the substitution.

  • $\begingroup$ I actually managed to solve the problem too few hours ago but I did it with using Laurent series of the logarithm. But it seems that piecewise function is easier way to get the desired result. $\endgroup$ – Harto Saarinen Aug 2 '14 at 17:14
  • $\begingroup$ +1. It's a fine answer. I was trying to do it but I don't understand the OP notation $\left[\ldots,\ldots\right]$ with $p = {6{\rm i} \over 5}$. $\endgroup$ – Felix Marin Aug 2 '14 at 17:55
  • $\begingroup$ Thats not so commonly used but it just means that C is two line segments: from 1 to p and from p to -1. $\endgroup$ – Harto Saarinen Aug 2 '14 at 20:55

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