Evaluate Case of Fresnel Integral

[Deanmark]

Evaluate $\lim\limits_{{n}\to{\infty}} \int_{n}^{n+1} \cos^2(x^2) \,dx$
I've tried using the half angle identity and the taylor series on the remaining $1/2 + \cos(2x^2)$ to prove the value is $1/2$, but I am out of ideas.

 

[I like Serena]

Evaluate $\lim\limits_{{n}\to{\infty}} \int_{n}^{n+1} \cos^2(x^2) \,dx$
I've tried using the half angle identity and the taylor series on the remaining $1/2 + \cos(2x^2)$ to prove the value is $1/2$, but I am out of ideas.

Hi Deanmark! Welcome to MHB! (Smile)

Can't we use that $C(x)= \int_0^x \cos(t^2)dt \to \sqrt{\frac{\pi}{8}} $ when $x\to \infty$?
See wiki.So $\int_{n}^{n+1} \cos(x^2)dx = C(n+1)-C(n) \to \sqrt{\frac{\pi}{8}} - \sqrt{\frac{\pi}{8}} = 0$ when $n\to \infty$.

According to the wiki article, we can prove it with a contour integral of the function $e^{-t^2}$.

 

[Greg]

It seems Wolfram Alpha uses a different version of the Fresnel C integral. The limit, as given there, is $\frac12$.

 

[I like Serena]

It seems Wolfram Alpha uses a different version of the Fresnel C integral. The limit, as given there, is $\frac12$.

Indeed, Wiki and Wolfram Mathworld list slightly different definitions.
To be fair, that's not uncommon with such special integral functions. We see the same thing with for instance Fourier transforms, where different sources specify different normalization constants.

Either way, the way I mentioned C(x) above, effectively includes which definition I'm using, eliminating the ambiguity.

 

[Opalg]

It seems Wolfram Alpha uses a different version of the Fresnel C integral. The limit, as given there, is $\frac12$.

The $\frac12$ comes from the fact that you want the integral of $\cos^2(x^2)$. As mentioned in the OP, $\cos^2(x^2) = \frac12(1 + \cos(2x^2))$. The integral of the constant $\frac12$ from $n$ to $n+1$ gives the $\frac12$ in the answer.

 

[Deanmark]

The question is, how can we prove $\lim_{{n}\to{n+1}}\int_{n}^{n+1}cos(2x^2)/2 \,dx = 0$? The best I have been able to come up with is the fact that $\lim_{{n}\to{n+1}}\ \int_{0}^{n+1}cos(2x^2)/2 \,dx - \int_{0}^{n}cos(2x^2)/2 \,dx$ = $\int_{n}^{n+1}cos(2x^2)/2 \,dx$ and each function is periodic and even so there must be cancelation, but it feels like I am making leaps.

 

[I like Serena]

The $\frac12$ comes from the fact that you want the integral of $\cos^2(x^2)$. As mentioned in the OP, $\cos^2(x^2) = \frac12(1 + \cos(2x^2))$. The integral of the constant $\frac12$ from $n$ to $n+1$ gives the $\frac12$ in the answer.

That's a different constant of $\frac 12$.
This one comes from Wolfram Mathworld that defines $C(u)=\int_0^u \cos\frac 12\pi x^2\, dx$ which approaches to the normalized $\frac 12$ instead of $\sqrt{\frac \pi 8}$.

The other $\frac 12$ comes indeed from the $\cos^2$ in the problem statement.

The question is, how can we prove $\lim_{{n}\to{n+1}}\int_{n}^{n+1}cos(2x^2)/2 \,dx = 0$? The best I have been able to come up with is the fact that $\lim_{{n}\to{n+1}}\ \int_{0}^{n+1}cos(2x^2)/2 \,dx - \int_{0}^{n}cos(2x^2)/2 \,dx$ = $\int_{n}^{n+1}cos(2x^2)/2 \,dx$ and each function is periodic and even so there must be cancelation, but it feels like I am making leaps.

How about:
\begin{aligned}
\lim_{{n}\to\infty} \int_{n}^{n+1}\cos(2x^2)/2 \,dx
&= \lim_{{n}\to\infty} \int_{0}^{n+1}\cos(2x^2)/2 \,dx - \int_{0}^{n}\cos(2x^2)/2 \,dx \\
&= \lim_{{n}\to\infty} \tilde C(n+1) - \tilde C(n) \\
&= D - D \\
&= 0
\end{aligned}
where $\tilde C$(n) is the variant of the Fresnel C function we have, since we have a couple of extra constants in there, and where $D$ is whatever it converges to at infinity.

 

[Deanmark]

That's a different constant of $\frac 12$.
This one comes from Wolfram Mathworld that defines $C(u)=\int_0^u \cos^2\frac 12\pi x^2\, dx$ which approaches to the normalized $\frac 12$ instead of $\sqrt{\frac \pi 8}$.

The other $\frac 12$ comes indeed from the $\cos^2$ in the problem statement.How about:
\begin{aligned}
\lim_{{n}\to\infty} \int_{n}^{n+1}\cos(2x^2)/2 \,dx
&= \lim_{{n}\to\infty} \int_{0}^{n+1}\cos(2x^2)/2 \,dx - \int_{0}^{n}\cos(2x^2)/2 \,dx \\
&= \lim_{{n}\to\infty} \tilde C(n+1) - \tilde C(n) \\
&= D - D \\
&= 0
\end{aligned}
where $\tilde C$(n) is the variant of the Fresnel C function we have, since we have a couple of extra constants in there, and where $D$ is whatever it converges to at infinity.

I think that is the best way to go. If we (I) can show it converges to the same constant, then the problem is done.

 

[I like Serena]

I think that is the best way to go. If we (I) can show it converges to the same constant, then the problem is done.

What we need is that $\tilde C(n)$ converges, and the Fresnel C integral $C(n)$ does.
If it does, both $\tilde C(n+1)$ and $\tilde C(n)$ converge to the same constant. (Nerd)

 

[Greg]

Hmm...

W|A

 

[Euge]

Evaluate $\lim\limits_{{n}\to{\infty}} \int_{n}^{n+1} \cos^2(x^2) \,dx$
I've tried using the half angle identity and the taylor series on the remaining $1/2 + \cos(2x^2)$ to prove the value is $1/2$, but I am out of ideas.

Using the u-substitution $u = x^2$, write the integral as $\displaystyle \int_{n^2}^{(n+1)^2} \frac{\cos^2 u}{2\sqrt{u}}\, du$. By the trig identity $\cos^2 u = \dfrac{1 + \cos 2u}{2}$, $$\int_{n^2}^{(n+1)^2} \frac{\cos^2 u}{2\sqrt{u}}\, du = \int_{n^2}^{(n+1)^2} \frac{1 + \cos 2u}{4\sqrt{u}}\, du$$ Now $\displaystyle \int_{n^2}^{(n+1)^2} \frac{1}{4\sqrt{u}}\, du = \frac{1}{2}\sqrt{(n+1)^2} - \frac{1}{2}\sqrt{n^2} = \frac{1}{2}$ and $$\int_{n^2}^{(n+1)^2} \frac{\cos 2u}{4\sqrt{u}}\, du = \frac{\sin 2u}{8\sqrt{u}}\bigg|_{u\, =\, n^2}^{(n+1)^2} + \int_{n^2}^{(n+1)^2}\frac{\sin 2u}{16u^{3/2}}\, du\tag{*}\label{eq1}$$ by integration by parts. Since $\lvert \sin 2u\rvert \le 1$, it follows that the right-hand side of \eqref{eq1} is bounded by $\dfrac{C}{n}$ where $C$ is some constant. Thus $\displaystyle \lim_{n\to \infty} \int_{n^2}^{(n+1)^2} \frac{\cos 2u}{4\sqrt{u}}\, du = 0$; consequently $\displaystyle \int_{n^2}^{(n+1)^2} \frac{1 + \cos 2u}{4\sqrt{u}}\, du \to \frac{1}{2}$ as $n\to \infty$, i.e., $\displaystyle \lim_{n\to \infty}\int_{n}^{n+1} \cos^2(x^2)\, dx = \frac{1}{2}$.

 

[Greg]

Hi Euge

Using the u-substitution $u = x^2$, write the integral as $\displaystyle \int_{n^2}^{(n+1)^2} \frac{\cos^2 u}{2\sqrt{u}}\, du$. By the trig identity $\cos^2 u = \dfrac{1 + \cos 2u}{2}$, $$\int_{n^2}^{(n+1)^2} \frac{\cos^2 u}{2\sqrt{u}}\, du = \int_{n^2}^{(n+1)^2} \frac{1 + \cos 2u}{4\sqrt{u}}\, du$$

This much I understand.

Now $\displaystyle \int_{n^2}^{(n+1)^2} \frac{1}{4\sqrt{u}}\, du= \frac{1}{2}\sqrt{(n+1)^2} - \frac{1}{2}\sqrt{n^2} = \frac{1}{2}$

I don't understand how the result of $\frac12$ is achieved. Can you clarify, please?

 

[Euge]

I don't understand how the result of $\frac12$ is achieved. Can you clarify, please?

An antiderivative for $\dfrac{1}{4\sqrt{u}}$ is $\dfrac{1}{2}\sqrt{u}$, so $$\int_{n^2}^{(n+1)^2} \frac{du}{4\sqrt{u}}\, du = \frac{1}{2}\sqrt{u}\bigg|_{u\, =\, n^2}^{(n+1)^2} = \frac{1}{2}\sqrt{(n+1)^2} - \frac{1}{2}\sqrt{n^2} = \frac{1}{2}(n+1) - \frac{1}{2}n = \frac{1}{2}$$

 

[Greg]

Ooops. Thanks!