Show that $\lim_{x \to \infty} \int_0 ^x \cos (t^3 + t) \,dt$ exists

Question

I have a tricky comprehensive exam problem that I've been wrestling with: it is exactly the problem in the title reproduced here:

Show that $$\lim_{x \to \infty} \int_0 ^x \cos (t^3 + t) \,dt$$ exists.

If we plot the function $f(t) = \cos (t^3 + t)$ then it's "obvious" by inspection that the given limit exists: the periods for cosine are decreasing to zero. Even so, I don't really see an obvious candidate for what the limit might be (I believe it is positive.)

My first step was to write down just exactly what it means that the limit exists. The limit exists if there exists some number $L < \infty$ with the following property: for any $\varepsilon >0$ there exists some $y >0$ such that

$$ x >y \ \ \implies \left| \int_0 ^x \cos (t^3 + t) \,dt - L \right| < \varepsilon $$

This is fine and good, but I immediately run in to the issue of not having a limit candidate $L$.

There's a problem in (baby) Rudin (Ch 6 Exc 13) which looks quite similar (in fact it was also given as a past comprehensive exam question) which is (in effect):

Show that $$ \lim_{x \to \infty} \int_0 ^x \sin (t^2) \,dt $$ exists.

I'll note that Rudin walks through the solution to the above in 4 or 5 steps no one of which is obvious or trivial (at least not to me)...

Thus my question: is there some easy (read: easy enough that it would be reasonable to expect the problem to be solved in a comprehensive exam setting) way to solve one or both of these problems?

Answer 1

By Dirichlet's test, if $f(x)$ is differentiable, bounded and decreasing to $0$ on $\mathbb{R}^+$, then the limit $$ \lim_{M\to +\infty}\int_{0}^{M}\cos(x)f(x)\,dx $$ is finite. Now $g(t)=t+t^3$ is increasing, differentiable and unbounded on $\mathbb{R}^+$, together with its inverse function $h(t)$. In a right neighbourhood of the origin $h(t)\sim t$, while in a left neighbourhood of $+\infty$ we have $h(t)\sim\sqrt[3]{t}$. By enforcing the substitution $t=h(x)$ we have

$$ \int_{0}^{M}\cos(t+t^3)\,dt = \int_{0}^{g(M)}\cos(x) h'(x)\,dx $$ and we may apply the previous lemma since $h$ is concave on $(1,+\infty)$.

---

With minor adjustements this approach proves the following useful Lemma:

If $f(x)/x$ is a differentiable, increasing and unbounded function on $\mathbb{R}^+$, the limit $\lim_{M\to +\infty}\int_{0}^{M}\exp\left(i f(x)\right)\,dx$ is finite.

---

About Fresnel's integral, $$ \int_{0}^{M}\sin(x^2)\,dx = \int_{0}^{M^2}\frac{\sin(x)}{2\sqrt{x}}\,dx\stackrel{\text{IBP}}{=}\left[\frac{1-\cos x}{2\sqrt{x}}\right]_{0}^{M^2}+\int_{0}^{M^2}\frac{1-\cos x}{4x\sqrt{x}}\,dx $$ where the first term of the RHS tends to zero as $M\to +\infty$. The last integral is the integral of a positive function, where $$ \int_{0}^{1}\frac{1-\cos x}{4x\sqrt{x}}\,dx \leq \int_{0}^{1}\frac{\frac{x^2}{2}}{4x\sqrt{x}}\,dx=\frac{1}{12} $$ and for any $M\geq 1$ $$ \int_{1}^{M^2}\frac{1-\cos x}{4x\sqrt{x}}\,dx \leq \int_{1}^{M^2}\frac{2}{4x\sqrt{x}}\,dx \leq \int_{1}^{+\infty}\frac{dx}{2x\sqrt{x}} = 1 $$ so the limit $\lim_{M\to +\infty}\int_{0}^{M}\sin(x^2)\,dx$ is positive, finite and less than $\frac{13}{12}$. Its actual value is $\frac{1}{2}\sqrt{\frac{\pi}{2}}$, as can be shown through the (inverse) Laplace transform or contour integration.

---

The actual value of $\lim_{M\to +\infty}\int_{0}^{M}\cos(t+t^3)\,dt$ is related to Airy's differential equation $f''(x)=x\cdot f(x)$ and it is given by $$ \frac{\pi}{\sqrt[3]{3}}\,\text{Ai}\left(\frac{1}{\sqrt[3]{3}}\right)=\frac{1}{3}\sum_{k=0}^\infty {1\over k!}\Gamma\left({k+1\over 3}\right)\sin\left({2\pi (k+1)\over 3}\right)\approx \sqrt{2}-1.$$

Answer 2

Hint.

$$ t^3+t = \frac{\pi}{2}+k\pi $$

always have a real root $t_k$ for each $k$ So the integral can be considered as

$$ \sum_{k = 0}^n\int_{t_k}^{t_{k+1}}\cos(t^3-t)dt $$

which is an alternating series and also $t_{k-1}-t_k < t_k - t_{k-1}$

so for $t > 1$

$$ \int_{t_{k+1}}^{t_{k+2}}| \cos(t^3-t)| dt < \int_{t_{k}}^{t_{k+1}}|\cos(t^3-t)|dt $$

giving an alternate series absolutely convergent.