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.


3 Answers 3


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

\begin{equation} \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) \end{equation}

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

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

  • 1
    $\begingroup$ 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.. $\endgroup$
    – mvggz
    Oct 20, 2014 at 14:03
  • $\begingroup$ @mvggz That's why I stated in my answer a brutal way. Besides, only that approach crossed to mind at that time (͡• ͜ʖ ͡•) $\endgroup$ Oct 20, 2014 at 14:26
  • $\begingroup$ 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.. $\endgroup$
    – mvggz
    Oct 20, 2014 at 14:33
  • $\begingroup$ @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... ≧◠‿◠≦✌ $\endgroup$ Oct 20, 2014 at 14:37
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    $\begingroup$ Your welcome. I have to say, you're very good at drawing faces with symbols :) $\endgroup$
    – mvggz
    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

  • $\begingroup$ 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 $\endgroup$
    – mvggz
    Oct 20, 2014 at 14:18

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