9
$\begingroup$

I am supposed, in an exercise, to calculate the above integral by integrating $f(z) = e^{-z^{2}}$ on the following countor: enter image description here

I began by separating the path $\gamma$ into three paths (obvious from the picture), and parametrizing each as follows:

$\gamma_{1} : [0, R] \rightarrow \mathbb{C}$ with $\gamma_{1}(t) = t$

$\gamma_{2} : [0, \frac{\pi}{4}] \rightarrow \mathbb{C}$ with $\gamma_{2}(t) = Re^{it}$

$\gamma_{3} : [0, \frac{\sqrt{2}R}{2}] \rightarrow \mathbb{C}$ with $\gamma_{3}^{-}(t) = t + it$ (with reverse orientation).

Then we can say that $\displaystyle\int_{\gamma} f(z) dz = \displaystyle\int_{\gamma_{1}} f(z) dz + \displaystyle\int_{\gamma_{2}} f(z) dz - \displaystyle\int_{\gamma_{3}^{-}} f(z) dz = 0$ since the path is closed.

Now $\displaystyle\int_{\gamma_{1}} f(z) dz = \displaystyle\int\limits_{0}^{R} e^{-t^{2}} dt$. We also get $\displaystyle\int_{\gamma_{3}^{-}} f(z) dz = -(i + 1) \displaystyle\int\limits_{0}^{\frac{\sqrt{2}R}{2}}e^{-2it^{2}} dt$. After playing around with sine and cosine a bunch to evaluate that last integral, I get:

$$0 = \int\limits_{0}^{R} e^{-t^{2}} dt + \int\limits_{\gamma_{2}} f(z) dz - \frac{i + 1}{\sqrt{2}} \int\limits_{0}^{R} \cos(u^{2}) du + \frac{i - 1}{\sqrt{2}} \int\limits_{0}^{R} \sin(u^{2}) du$$

I could not evaluate the integral along the second path, but I thought it might tend to 0 as $R \rightarrow \infty$. Then taking limits and equating real parts we get

$$\frac{\sqrt{2 \pi}}{2} = \displaystyle\int\limits_{0}^{\infty} \sin(u^{2}) du + \displaystyle\int\limits_{0}^{\infty} \cos(u^{2}) du$$

If I could argue that the integrals are equal, I would have my result.. But how do I?

So I need to justify two things: why the integral along $\gamma_{2}$ tends to zero and why are the last two integrals equal.

$\endgroup$
  • 2
    $\begingroup$ Just as a comment $$\int\limits_0^\infty {\sin \left( {a{x^2}} \right)\cos \left( {2bx} \right)dx} = \sqrt {\frac{\pi }{{8a}}} \left( {\cos \frac{{{b^2}}}{a} - \sin \frac{{{b^2}}}{a}} \right)$$ $$\int\limits_0^\infty {\cos \left( {a{x^2}} \right)\cos \left( {2bx} \right)dx} = \sqrt {\frac{\pi }{{8a}}} \left( {\cos \frac{{{b^2}}}{a} + \sin \frac{{{b^2}}}{a}} \right)$$ $\endgroup$ – Pedro Tamaroff Apr 4 '12 at 2:41
1
$\begingroup$

Your parametrisation of the third integral is rather complicated. Why not just write $\gamma_3:[0,R]\rightarrow \mathbb{C}$, $t\mapsto -e^{\pi i/4}t$. Then the integral becomes $$ \int_R^0 e^{-e^{\pi i/2}t^2}e^{\pi i/4}dt = \int_R^0 e^{-it^2}e^{\pi i/4}dt = e^{\pi i/4} \int_R^0 \cos t^2 - i \sin t^2 dt. $$ I am sure you can take it from there.

As for bounding the integral $\int_{\gamma_2}e^{-z^2}dz$, the length of the contour grows linearly with $R$. How fast does the maximum of the integrand decay? It's the standard approach, using the fact that $$ \left|\int_\gamma f(z) dz\right|\leq \sup\{|f(z)|: z \in \text{ image of }\gamma\}\cdot \text{length of }\gamma. $$

$\endgroup$
  • $\begingroup$ Just use your inequality cannot solve the problem! $\endgroup$ – Lao-tzu Jan 12 '15 at 5:14
  • $\begingroup$ I think the best way to obtain what you want is to use the Lebesgue dominated theorem: as R tends to infinity, the integrand tends to 0 and is less than a constant which is integrable on a finite interval. $\endgroup$ – Lao-tzu Jan 12 '15 at 5:41
  • $\begingroup$ Sorry, not a constant, but still a integrable function on the closed interval. $\endgroup$ – Lao-tzu Jan 12 '15 at 7:29
1
$\begingroup$

$$\int_0^R e^{-t^2}dt= \frac{\sqrt{\pi}}{2}+o(1). \tag{1}$$

$$t\in\left[0,\frac{\pi}{4}\right] \implies \operatorname{Re}(Re^{it})\ge \frac{R}{\sqrt{2}} \implies \left|\int_{\gamma_2}e^{-z^2}dz\right|\le R\frac{\pi}{4}e^{-R^{\,2}/2}\to0 \tag{2}$$

$$\alpha=\int_0^M e^{it^2}dt \implies \int_0^M \sin(t^2)dt=\frac{\alpha+\overline{\alpha}}{2}. \tag{3}$$

Try using these deductions for $\gamma_1$, $\gamma_2$ and $\gamma_3$ respectively.

$\endgroup$
1
$\begingroup$

An easy way to evaluate $\int_{0}^{\infty}\sin(x^{2})dx$

$$\int_0^{\infty}e^{-ax^2}dx=\frac{\sqrt{\pi}}{2\sqrt{a}}$$ Now replace $a\rightarrow ia$

$$ \int\limits_0^\infty \cos \left( {a{x^2}}\right)dx-i \int\limits_0^\infty \sin \left( {a{x^2}}\right)dx= \frac{\sqrt{\pi}}{2\sqrt{a}\sqrt{i}} $$ But

$$\frac{1}{\sqrt{i}}= \frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}} $$ So

$$ \int\limits_0^\infty \cos \left( {a{x^2}}\right)dx=\int\limits_0^\infty \sin \left( {a{x^2}}\right)dx= \frac{\sqrt{\pi}}{2\sqrt{2a}} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.