# Thread: [SOLVED] residue theorem (definite real integrals)

1. ## [SOLVED] residue theorem (definite real integrals)

Hi everyone,

I didn't know where I should post my question as I'm new here; I hope this is the right place. Anyway, my question is:

I need to show that the following integral:

$\displaystyle \int_0^{\infty} {\cos(t^2)}. dt$

is equal to = $\displaystyle \frac{\sqrt\pi}{2\sqrt2}$

we're also given (i.e. can use)

$\displaystyle \int_0^{\infty} {\exp(-u^2)}. du = \frac{\sqrt\pi}{2}$

What is the best contour?

I think the residue will be zero as the function is analytic everywhere but the problem is how to prove that the contours used tend to zero?

2. Originally Posted by tbm206
Hi everyone,

I didn't know where I should post my question as I'm new here; I hope this is the right place. Anyway, my question is:

I need to show that the following integral:

$\displaystyle \int_0^{\infty} {\cos(t^2)}. dt$

is equal to = $\displaystyle \frac{\sqrt\pi}{2\sqrt2}$

we're also given (i.e. can use)

$\displaystyle \int_0^{\infty} {\exp(-u^2)}. du = \frac{\sqrt\pi}{2}$

What is the best contour?

I think the residue will be zero as the function is analytic everywhere but the problem is how to prove that the contours used tend to zero?
Hi,

Notice that your integral is the real part of $\displaystyle \int_0^\infty e^{it^2}dt$, so it suffices to compute this one.

Now, let $\displaystyle f(z)=e^{i z^2}$.

The integral you want is obtained by integrating $\displaystyle f$ along the real axis: $\displaystyle z=t$.

The integral you know arises when $\displaystyle z^2=i t^2$, which means (for instance) $\displaystyle z= e^{i\frac{\pi}{4}}t$ (I took a square root of $\displaystyle i=e^{i\frac{\pi}{2}}$).

So you should prove that integrating $\displaystyle f$ along $\displaystyle [0,+\infty)$ gives the same result as integrating $\displaystyle f$ along $\displaystyle e^{i\pi/4}[0,+\infty)$ (this is the bisector of the north-east quarter plane).

Given $\displaystyle R>0$, you could consider the contour that goes from 0 to $\displaystyle R$ on $\displaystyle \mathbb{R}$, then follows a piece of circle into $\displaystyle Re^{i\pi/4}$, and goes back to 0 along the radius "$\displaystyle [0,e^{i\pi/4}]$". You know what the integral along this contour is equal to (as you said, there is no singularity). Then the idea is to make $\displaystyle R\to\infty$ and prove that the integration on the piece of circle tends to 0. This would give you the equality you need.

Write down what is the integral of $\displaystyle f$ along the three different parts of the contour (the two radii and the piece of arc) to try to complete the proof. If you're still stuck, tell us what you found.

3. You might like to know that this is called a Fresnel integral.

4. Here are the steps that Laurent mentions in his post.

Let $\displaystyle \Gamma$ be the contour from $\displaystyle 0$ to $\displaystyle R$ then transversed counterclockwise to $\displaystyle Re^{i\pi/4}$ and then back (in a straight line) down to $\displaystyle 0$.

Let $\displaystyle f(z) = e^{iz^2}$.

Thus, $\displaystyle \oint_{\Gamma} f(z) dz = 0$ by Cauchy's theorem but this implies:
$\displaystyle \int_0^R e^{ix^2} dx + \int_0^{\pi/4} R\cdot if(Re^{i\theta})e^{i\theta} d\theta - \int_0^R e^{i\pi/4} f(it^2e^{i\pi/2}) dt = 0$

We want to see what happens when we take the limit $\displaystyle R\to \infty$.
The idea is to show that the middle integral goes to zero.

Notice that $\displaystyle |R\cdot if(Re^{i\theta})| = |R\cdot \exp(R^2\sin 2\theta - iR^2\cos \theta)| = Re^{-R^2\sin 2\theta}$

Now we need an inequality. Draw the graph of $\displaystyle y=\sin \theta$ on $\displaystyle [0,\tfrac{\pi}{2}]$ and notice that the line joining $\displaystyle (0,0)$ and $\displaystyle (\tfrac{\pi}{2},1)$ lies below the sine curve. Therefore, we see that $\displaystyle \sin x \geq \tfrac{2}{\pi}x$ for $\displaystyle 0\leq x\leq \tfrac{\pi}{2}$. And so it follows that $\displaystyle Re^{-R^2\sin 2\theta} \leq Re^{-4R^2\theta /\pi}$.

Thus, $\displaystyle \left| \int_0^{\pi/4} R\cdot if(Re^{i\theta})e^{i\theta} d\theta \right| \leq R \int_0^{\pi/4} e^{-4R^2\theta/\pi}d\theta = \frac{\pi(1 - e^{-R^2})}{4R}\to 0 \text{ as }R\to \infty$

Thus, the middle integral is zero in the limit,
$\displaystyle \int_0^{\infty} e^{ix^2} dx = e^{i\pi/4} \int_0^{\infty} e^{-t^2} dt$

Finish it from there.

5. Thanks guys!

As a matter of fact, I knew that I have to use the "pizza-slice" contour but the difficulty rose when I tried to prove that the second integral goes to zero. Actually, even after reading Mr. ThePerfectHacker's solution I can't understand the bit:

$\displaystyle |R\cdot jf(Re^{j\theta})| = |R\cdot \exp(R^2\sin 2\theta - jR^2\cos \theta)| = Re^{-R^2\sin 2\theta}$

The rest is clear

6. Originally Posted by tbm206
Thanks guys!

As a matter of fact, I knew that I have to use the "pizza-slice" contour but the difficulty rose when I tried to prove that the second integral goes to zero. Actually, even after reading Mr. ThePerfectHacker's solution I can't understand the bit:

$\displaystyle |R\cdot jf(Re^{j\theta})| = |R\cdot \exp(R^2\sin 2\theta - jR^2\cos \theta)| = Re^{-R^2\sin 2\theta}$

The rest is clear

You need to remember that if $\displaystyle \alpha\in \mathbb{R}$ then $\displaystyle |e^{i\alpha}| = 1$.
Remember that $\displaystyle f(z) = e^{iz^2}$ therefore $\displaystyle |R\cdot i\cdot f(Re^{i\theta})| = R|e^{i(Re^{i\theta})^2}| = R|e^{-R^2\sin2\theta + i R^2\cos 2\theta}|$
We get, $\displaystyle Re^{-R^2\sin 2\theta}|e^{iR^2\cos 2\theta}| = Re^{-R^2\sin 2\theta}$.
7. Oh!! I see; I can't believe how I missed that $\displaystyle |e^{j\alpha}| = 1$;