Hi: I'm reading some notes I found on complex analysis on the internet. In the example, they prove that

$$\int_{0}^{\infty} \cos(x^2) = \int_{0}^{\infty} \sin(x^2) = \frac{\sqrt{2\pi}}{4}.$$

I understand a lot of the proof so I don't think that I need to write down all the steps. I will just provide what I think is necessary for readers to understand my lack of understanding.

The example starts off with

"We shall consider the analytic function $f(z) = e^{(iz^2)}$ over a contour C of the form of a circular wedge in the first quadrant. Then contour has three edges:

(1) a line segment of length R from the $0$ to the point $A$.=,

(2) from $A$ to $B$ over a circular arc of angle $\frac{\pi}{4}$ (so $B = R e^{i \pi/4})$ and

(3) from $B$ to $0$.

That make sense. Then the author breaks the integral into three pieces so that

$$\int_{C} f(z) dz = \int_{OA} f(z) dz + \int_{AB} f(z) dz + \int_{BO} f(z) dz = 0.$$

by cauchy's theorem. That makes sense too.

But then, when evaluating the third integral, the author has

$$\begin{align} I_{3}(R) &= \int_{BO} e^{(i z^2)} dz \\&= \int_{R}^{0} e^{-r^2} e^{i \pi/4} dr \\&= -e^{i \pi/4} \int_{0}^{R} e^{-r^2} dr. \end{align}$$

Then the author explains

" for $I_{3}(R) $ we used $ z = r e^{i \pi/4}, 0 \le r \le R; z^2 = r^2e^{i \pi/4}; dz = e^{i \pi/4} dr $ ".

So, given what was used for $I_{3}(R)$ , I don't see how that integral is obtained. I must be missing something but could someone spell out how the negative comes about ? In fact, I don't even see where the $e^{i \pi/4}$ part of $z^2$ went ? Basically, I don't follow the integral construction at all.

Also, my second confusion ( definitely there's a typo somewhere causing this confusion ) occurs immediately after above where the author states:

" Next observe that: $\displaystyle I_{3}(R) = -e^{i\pi/4} \int_{0}^{R} e^{-r^2} dr = - \frac{\sqrt{2}}{2}(1 + i) \int_{0}^{R} e^{-r^2} dr. $ As $R \longrightarrow \infty$, we find : $\lim_ {R\to\infty} I_{3}(R) = - \frac{\sqrt{2 \pi}}{2} (1 + i).$ "

Is the above correct ( I don't know how to evaluate the integral ) because later on in the proof, he states that

$ - I_{3}(R) = \frac{\sqrt{2 \pi}}{4} (1 + i).$ So clearly there has to be a typo in one of the statements about what $I_{3}(R)$ is equal to.

Thank you very much for clearing up my two confusions.


On second confusion: from http://en.wikipedia.org/wiki/Gaussian_integral

$$\int_{-\infty}^{+\infty} e^{-x^2}\,dx = \sqrt{\pi}$$


$$\lim_{R \to \infty} \int_0^R e^{-r^2} = \frac{\sqrt{\pi}}{2}$$ therefore $\lim_{R \to \infty} I_3(R) = \lim_{R \to \infty} (-\frac{\sqrt{2}}{2}) (1 + i) \int_0^R e^{-r^2} = -\frac{\sqrt{2}}{2} (\lim_{R \to \infty} \int_0^R e^{-r^2} + i \lim_{R \to \infty} \int_0^R e^{-r^2}) = -\frac{\sqrt{2}}{2} (\frac{\sqrt{\pi}}{2} + i\frac{\sqrt{\pi}}{2}) = - \frac{\sqrt{2 \pi}}{4} (1 + i)$

  • $\begingroup$ hi richard. nice explanation. now if I can just figure out how they got the initial integral, I'll be all set. thanks a lot. $\endgroup$ – mark leeds Jul 31 '14 at 15:47
  • $\begingroup$ on the second confusion, I think they got the negative by flipping the upper and limits which is fine. but then when $z^2$ is substituted for, it seems that the $\frac{i \pi}{4}$ just becomes the multiplier of the $e^{-r^2}$ which I don't get. thanks again for any clarity there. $\endgroup$ – mark leeds Jul 31 '14 at 17:01
  • $\begingroup$ actually, studying it more, my guess is that the $e^{\frac{i \pi}{4}}$ is mostly likely coming from the dz and the other one is dissapearing but I don't see how. thanks. $\endgroup$ – mark leeds Jul 31 '14 at 17:06

I decided to answer the question on the complex integration tag so the following answer is attributed to @Per Manne. I will check off the answer.


There seems to be a misprint in the part which says what $z^2$ is supposed to be. Since $z=r{\rm e}^{{\rm i}\pi/4}$ on $BO$, you get that $$z^2=r^2{\rm e}^{{\rm i}\pi/2}={\rm i}r^2$$ and hence $f(z)={\rm e}^{-r^2}$.

Putting things together, just substitute ${\rm d}z={\rm e}^{{\rm i}\pi/4}{\rm d}r$ and remember that along the segment $BO$, the parameter $r$ will start at $R$ and decrease to $0$. The minus sign in the last expression for $I_3(R)$ comes from switching the direction and letting $r$ run from $0$ to $R$ instead.


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