# Proving convergence of $\int \limits_0^{\infty} \cos\left(x^2\right) dx$

How would one prove the convergence of $$\int_0^{\infty} \cos\left(x^2\right) \,\mathrm dx$$

I tried using the integral test for convergence by noting that making the substitution $u = x^2$ means you can write the integral in the form:

$$\int_0^{\infty} \frac{1}{2\sqrt{u}}\cos(u) \space \mathrm{du}$$

I'm not sure what to do here - I though of rewriting the cosine as a Taylor series, and then applying the integral test, but I don't think that would work.

• See the evaluation of Fresnel integrals. Oct 19, 2014 at 11:29
• possible duplicate of Prove: $\int_{0}^{\infty} \sin (x^2) dx$ converges. Oct 19, 2014 at 11:32
• @GitGud It's not immediately clear to me how to go from $\int \sin(x^2) dx$ to $\int \cos(x^2) dx$. I don't know if it's really a duplicate. Oct 19, 2014 at 11:53
• Oct 19, 2014 at 12:17
• @NajibIdrissi I retracted my vote. Oct 19, 2014 at 12:29

The convergence of such integral follows from the continuous analogue of the Dirichlet criterion for the convergence of a series.

Dirichlet criterion: If $\{a_n\}_{n\in\mathbb{N}}$ is a sequence with the property that the partial sums $\sum_{k=0}^n a_k$ are bounded, and $\{b_n\}_{n\in\mathbb{N}}$ is a decreasing sequence converging towards zero, the series $$\sum_{n=0}^{+\infty} a_n b_n$$ is convergent. The proof relies on summation by parts.

Dirichlet criterion (integral version): If $f(x)$ is a Riemann-integrable function with the property that $\int_{0}^{x}f(t)\,dt$ is bounded for any $x\geq 0$, and $g(x)$ is a continuous decreasing function over $\mathbb{R}^+$ such that $\lim_{x\to +\infty}g(x)=0$, then $f\cdot g$ is a Riemann-integrable function over $\mathbb{R}^+$.

The proof relies on integration by parts.

In order to apply the last criterion, it is sufficient to take $f(x)=\cos x$ and $g(x)=\frac{1}{2\sqrt{x}}$.

This is a brutal way to prove the convergence of the given integral. We have the following formula

$$\int_0^\infty \cos ax^2\cos 2bx\, dx=\frac{1}{2}\sqrt{\frac{\pi}{2a}}\left(\cos \frac{b^2}{a}+\sin\frac{b^2}{a}\right)$$

The detail proof can be seen here: Duo Fresnel-like integrals $(??)$

Therefore, setting $a=1$ and $b=0$, we get $$\int_0^\infty \cos x^2\, dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}=\sqrt{\frac{\pi}{8}}$$ Addendum of the proof in the cited link, we can use the fact $$\int_{-\infty}^\infty e^{-ax^2}\,dx=\sqrt{\frac{\pi}{a}}$$

• That doesn't show how you actually treat such a question, quoting very sophisticated formulas to deal with basic questions like this one is not a way of helping the OP. That means he should prove that first, which is quite more complicated, and only then will he be able to conclude. That's like using a hammer to kill a ladybug.. Oct 20, 2014 at 14:03
• @mvggz That's why I stated in my answer a brutal way. Besides, only that approach crossed to mind at that time (͡• ͜ʖ ͡•) Oct 20, 2014 at 14:26
• I apologize if I gave you the feeling I was judging or anything, it's just that I've always been taught not to overuse results when we coud avoid it. In fact, my teacher would grant me a zero whenever I would do that, because we should always be able to make the best out of what we had learned. For instance, I could have had this question on an exam, and was expected to answer quickly and using very little material. That's why I thought it was important.. Oct 20, 2014 at 14:33
• @mvggz No need to apologize. I was in hurry, you know I'm chasing reps here, lol. I'll take your advice as a consideration. Thanks... ≧◠‿◠≦✌ Oct 20, 2014 at 14:37
• Your welcome. I have to say, you're very good at drawing faces with symbols :) Oct 20, 2014 at 14:43

I can give you a fast way of treating this question:

First you notice that your function is continuous on $[0,+\inf[$

So your integral is defined on [0,1] for instance:

I(X) = $\int_{[0,X]} cos(t^2)dt = \int_{[0,1]} cos(t^2)dt + \int_{[1,X]} cos(t^2)dt = A + I_1(X)$

$I_1(X) = \int_{[1,X]} 2t*\frac{cos(t^2)}{2t}dt$ = $[\frac{sin(t^2)}{2t}]_1^X + \int_{[1,X]} \frac{sin(t^2)}{2t^2}dt$

The first term is a constant and a term that ->0 when x->inf. The second verifies:

$|g(t)| = |\frac{sin(t^2)}{2t^2}| \leq \frac{1}{2t^2}$ => g integrable on [1,+inf[ , because it is dominated by an integrable function on [1, +inf[

Hence $I_1(X)$ has a limit when X->inf, and so does your integral

• The idea is to get away from 0 before using a integration by part on the $t*cos(t^2)$, because zero does not fit with \frac{1}{t} in terms of integrability, and the reasoning collapses. But if you start the integral at 1 instead of zero the complication vanishes Oct 20, 2014 at 14:18