How to evaluate the limit: $\lim_{t \to \infty} \int_0^1 \frac{e^{it^2(1+y^2)}}{1+y^2} \, dy$

Question

I was evaluating the Fresnel integral: $$ \int_{-\infty}^{\infty} e^{ix^2}dx $$

After some calculations, I successfully evaluated it correctly. However, there is one problem that I don't know how to evaluate strictly and not just logically. I want to evaluate this limit:

$$ \lim_{{t \to \infty}} \int_{0}^{1} \frac{e^{it^2(1+y^2)}}{1+y^2} \, dy $$

So, I have done this:

Let $y={x\over t}$, then our limit will become:

$$ \lim_{{t \to \infty}} \int_{0}^{t} \frac{e^{it^2 \left(1+\frac{x^2}{t^2}\right)}}{1+\frac{x^2}{t^2}} \frac{1}{t} \, dx $$

Now, let's do something like this:

$$ \begin{align*} &\lim_{{t \to \infty}} \int_{0}^{t} \frac{e^{it^2(1+\frac{x^2}{t^2})}}{1+\frac{x^2}{t^2}} \frac{1}{t} \, dx \\ &= \lim_{{t \to \infty}} \left(\int_{0}^{t} \frac{\cos(t^2(1+\frac{x^2}{t^2}))}{1+\frac{x^2}{t^2}} \frac{1}{t} \, dx + \int_{0}^{t} \frac{i\sin(t^2(1+\frac{x^2}{t^2}))}{1+\frac{x^2}{t^2}} \frac{1}{t} \, dx\right) \end{align*} $$

Now, let's look at the first member and evaluate the limit for it:

$$ \lim_{{t \to \infty}} \int_{0}^{t} \frac{\cos(t^2(1+\frac{x^2}{t^2}))}{1+\frac{x^2}{t^2}} \frac{1}{t} \, dx $$

Instead of this integral, let's evaluate this kind of integral:

$$ \int_{0}^{t} \frac{\left(\cos(t^2(1+\frac{x^2}{t^2}))\right)^2}{(1+\frac{x^2}{t^2})^2} \frac{1}{t^2} \, dx $$

Now, let's use the Dominated Convergence Theorem:

$$ \left| \int_{0}^{t} \frac{\left(\cos(t^2(1+\frac{x^2}{t^2}))\right)^2}{(1+\frac{x^2}{t^2})^2} \frac{1}{t^2} \, dx \right| \leq \int_{0}^{t} \frac{1}{(1+\frac{x^2}{t^2})^2} \frac{1}{t^2} \, dx $$

As we evaluate the integral on the right side, it will give us:

$$ \int_{0}^{t} \frac{1}{(1+\frac{x^2}{t^2})^2} \frac{1}{t^2} \, dx = \frac{\pi+2}{8t} $$

So, we have:

$$ \left| \int_{0}^{t} \frac{\left(\cos(t^2(1+\frac{x^2}{t^2}))\right)^2}{(1+\frac{x^2}{t^2})^2} \frac{1}{t^2} \, dx \right| \leq \frac{\pi+2}{8t} $$

Now, let's take the limit on both sides, and we will get:

$$ \left| \lim_{{t \to \infty}} \int_{0}^{t} \frac{\left(\cos(t^2(1+\frac{x^2}{t^2}))\right)^2}{(1+\frac{x^2}{t^2})^2} \frac{1}{t^2} \, dx \right| \leq 0 $$

And from this:

$$ \lim_{{t \to \infty}} \int_{0}^{t} \frac{\left(\cos(t^2(1+\frac{x^2}{t^2}))\right)^2}{(1+\frac{x^2}{t^2})^2} \frac{1}{t^2} \, dx = 0 $$

And from this:

$$ \lim_{{t \to \infty}} \int_{{0}}^{{t}} \frac{{\cos(t^2(1+\frac{x^2}{t^2}))}}{{1+\frac{x^2}{t^2}}} \cdot \frac{1}{t} \, dx = 0 $$ With the same logic, the second member is also $0$. Therefore:

$$ \lim_{{t \to \infty}} \int_{0}^{t} \frac{e^{it^2(1+\frac{x^2}{t^2})}}{1+\frac{x^2}{t^2}} \frac{1}{t} \, dx = 0 $$

But I can't properly explain why it was right to say if the limit of the integral from the square of the integrand function is $0$, then the limit of the integral from the integrand function (without the square) is also $0$. Can someone please explain this to me mathematically and not just logically? I know the first equation, which we desire to evaluate, oscillates very fast and therefore is zero, but I want to show all this mathematically, and I don't want to use the Gamma function.

Basically, we said that if: $$ \lim_{{t \to \infty}} \int_{{0}}^{{t}} \left(f(x;t)\right)^2 \, dx = 0 $$ then: $$ \lim_{{t \to \infty}} \int_{{0}}^{{t}} f(x;t) \, dx = 0 $$

But it isn't true for every function. Then, what makes it work in this case?

Thanks to everyone in advance who will try it.

Answer 1

We consider $I(t)=\int_0^1 \frac{e^{it^2 x}}{\sqrt x(1+x)} \,dx$ . A typical way to solve such problems is to apply integration by part. But in this case, as @ Sangchul Lee mentioned, we cannot do this directly (we get terms diverging at $x=0$).

We can also apply complex integration. Let's consider a closed contour (a quarter-circle) in the complex plane; namely, introducing small $r>0$, we integrate along the following path $C$: $$r\to1 \,(\text{along the real axis} \,X)\to i \,(\text{along the arch of the circle of the radius 1}\,(C_1))$$ $$\to ir\, (\text{along the imaginary axis} \,Y)\to r\,(\text{along the arch of the circle of the radius r}\,(C_2))$$ The function $f(z)=\frac{e^{it^2 z}}{\sqrt z(1+z)}$ is analytical inside the contour; therefore, $\oint_C f(z)dz=0$. Denoting also $I_r=\int_r^1 \frac{e^{it^2 x}}{\sqrt x(1+x)} \,dx$, we get the following equation: $$I_r+\int_{C_1}\frac{e^{it^2 z}}{\sqrt z(1+z)}dz+\int_i^{ir}\frac{e^{it^2 z}}{\sqrt z(1+z)}dz+\int_{C_2}\frac{e^{it^2 z}}{\sqrt z(1+z)}dz=0\tag{0}$$ Next, we evaluate every integral $$\bigg|\int_{C_2}\frac{e^{it^2 z}}{\sqrt z(1+z)}dz\bigg|=\bigg|\int_{\pi/2}^0\frac{e^{it^2 re^{i\phi}}}{\sqrt re^{i\phi/2}(1+re^{i\phi})}ire^{i\phi}d\phi\bigg|$$ $$<\int_0^{\pi/2}\bigg|\frac{e^{it^2 re^{i\phi}}}{\sqrt re^{i\phi/2}(1+re^{i\phi})}ire^{i\phi}\bigg|d\phi=\sqrt r\int_0^{\pi/2}\frac{e^{-t^2r\sin\phi}}{\sqrt{1+r^2}}d\phi<\sqrt r\int_0^{\pi/2}d\phi\to 0$$ $$=\frac\pi2\,\sqrt r\,\to 0\,\,\text{at}\,\,r\to 0\tag{1}$$ $$I_r=\int_r^1 \frac{e^{it^2 x}}{\sqrt x(1+x)} \,dx=\int_0^1 \frac{e^{it^2 x}}{\sqrt x(1+x)} \,dx-\int_0^r \frac{e^{it^2 x}}{\sqrt x(1+x)} \,dx$$ where $$\bigg|\int_0^r \frac{e^{it^2 x}}{\sqrt x(1+x)} \,dx\bigg|<\int_0^r \bigg|\frac{e^{it^2 x}}{\sqrt x(1+x)}\bigg| \,dx=\int_0^r\frac{dr}{\sqrt r(1+r)}<\int_0^r\frac{dr}{\sqrt r}$$ $$=\frac{\sqrt r}2\to 0\,\,\text{at}\,\,r\to 0$$ It means that $$I_r\to I\,\,\text{at}\,\,r\to0\tag{2}$$ In the same way, $$\int_i^{ir}\frac{e^{it^2 z}}{\sqrt z(1+z)}dz\overset{z=e^{\pi i/2}x}{=}-\,e^{\pi i/4}\int_r^1\frac{e^{-t^2x}}{\sqrt x(1+ix)}dx$$ $$=-\,e^{\pi i/4}\int_0^1\frac{e^{-t^2x}}{\sqrt x(1+ix)}dx+\,e^{\pi i/4}\int_0^r\frac{e^{-t^2x}}{\sqrt x(1+ix)}dx$$ $$\int_i^{ir}\frac{e^{it^2 z}}{\sqrt z(1+z)}dz\to -\,e^{\pi i/4}\int_0^1\frac{e^{-t^2x}}{\sqrt x(1+ix)}dx\,\,\text{at}\,\,r\to 0\tag{3}$$ Leading $r\to 0$ and using (1), (2), (3) $$I=-\int_{C_1}\frac{e^{it^2 z}}{\sqrt z(1+z)}dz+\,e^{\pi i/4}\int_0^1\frac{e^{-t^2x}}{\sqrt x(1+ix)}dx\tag{4}$$ where $$-\int_{C_1}\frac{e^{it^2 z}}{\sqrt z(1+z)}dz=-\int_0^{\pi/2}\frac{e^{it^2e^{i\phi}}ie^{i\phi}}{e^{i\phi/2}(1+e^{i\phi})}d\phi$$ $$\overset{IBP}{=}\frac{i\,e^{-i\phi}e^{it^2e^{i\phi}}}{2t^2\cos\frac\phi2}\,\bigg|_{\phi=0}^{\phi=\frac\pi2}+\frac i{2t^2}\int_0^{\pi/2}\frac{e^{-i\phi}e^{it^2e^{i\phi}}}{\cos\frac\phi2}\left(i-\frac{\sin\frac\phi2}{2\cos^2\frac\phi2}\right)d\phi$$ $$=-\,\frac i{2t^2}+\frac{e^{-t^2}}{\sqrt 2t^2}+\frac i{2t^2}\int_0^{\pi/2}\frac{e^{-i\phi}e^{it^2e^{i\phi}}}{\cos\frac\phi2}\left(i-\frac{\sin\frac\phi2}{2\cos^2\frac\phi2}\right)d\phi$$ The remaining integral is regular and at least $\sim\frac1{t^2}$ (integrating by part again we can show that it $\sim\frac1{t^4}$) $$-\int_{C_1}\frac{e^{it^2 z}}{\sqrt z(1+z)}dz=O\left(\frac1{t^2}\right)\tag{5}$$

And the last integral $$e^{\pi i/4}\int_0^1\frac{e^{-t^2x}}{\sqrt x(1+ix)}dx\overset{s=t^2x}{=}\frac{e^{\pi i/4}}t\int_0^{t^2}\frac{e^{-s}}{\sqrt s(1+\frac{is}{t^2})}ds$$ $$=\frac{e^{\pi i/4}}t\int_0^\infty\frac{e^{-s}}{\sqrt s(1+\frac{is}{t^2})}ds-\frac{e^{\pi i/4}}t\int_{t^2}^\infty\frac{e^{-s}}{\sqrt s(1+\frac{is}{t^2})}ds$$ where $$\bigg|\frac{e^{\pi i/4}}t\int_{t^2}^\infty\frac{e^{-s}}{\sqrt s(1+\frac{is}{t^2})}ds\bigg|<\frac1t\int_{t^2}^\infty\frac{e^{-s}}tds=\frac{e^{-t^2}}{t^2}\tag{6}$$ and $$\frac{e^{\pi i/4}}t\int_0^\infty\frac{e^{-s}}{\sqrt s(1+\frac{is}{t^2})}ds=\frac{e^{\pi i/4}}t\int_0^\infty\frac{e^{-s}}{\sqrt s}ds+\frac{e^{\pi i/4}}t\int_0^\infty e^{-s}\left(\frac1{\sqrt s(1+\frac{is}{t^2})}-\frac1{\sqrt s}\right)ds=$$ $$=\frac{e^{\pi i/4}}t\sqrt\pi+\frac{e^{-\pi i/4}}{t^3}\int_0^\infty e^{-s}\frac{\sqrt s}{1+\frac{is}{t^2}}ds=\frac{e^{\pi i/4}\sqrt \pi}t+O\left(\frac1{t^3}\right)\tag{7}$$ In fact, repeating the procedure we can show that $(7)=\frac{e^{\pi i/4}\sqrt\pi}t+\frac{e^{-\pi i/4}\sqrt\pi}{2t^3}+O\left(\frac1{t^5}\right)$

Putting (5), (6) and (7) into (4) $$I=\frac{e^{\pi i/4}\sqrt \pi}t+O\left(\frac1{t^2}\right)$$ The limit follows.

Answer 2

New Answer. Since OP seems not aware of Lebesgue theory of integrals, let me switch to a more elementary means.

Key Idea. The basic idea is to note that

$$ \int y e^{it^2y^2} \, \mathrm{d}y = \frac{e^{it^2y^2}}{2it^2} $$

is bounded by $t^{-2}$, which vanishes as $t \to \infty$. So, if we apply integration by parts to OP's integral, we will end up with something that decays as $t \to \infty$, hinting that the limit is zero. Due to some technicality involved, however, I will implement this idea only indirectly.

Evaluation. We begin by noting that

$$ \frac{1}{1+y^2} = y \biggl( \frac{1}{2} + \int_{y}^{1} \frac{1+3x^2}{x^2(1+x^2)^2} \, \mathrm{d}x \biggr). $$

Now let $I(t) = \int_{0}^{1} \frac{e^{it^2(1+y^2)}}{1+y^2} \, \mathrm{d}y$ denote OP's integral. Plugging the above identity to $I(t)$ and switching the order of integration using Fubini's theorem,

\begin{align*} I(t) &= e^{it^2} \int_{0}^{1} e^{it^2y^2} y \biggl( \frac{1}{2} + \int_{y}^{1} \frac{1+3x^2}{x^2(1+x^2)^2} \, \mathrm{d}x \biggr) \, \mathrm{d}y \\ &= e^{it^2} \biggl( \frac{1}{2}\biggl[ \int_{0}^{1} e^{it^2y^2} y \, \mathrm{d}y \biggr] + \int_{0}^{1} \frac{1+3x^2}{x^2(1+x^2)^2} \biggl[ \int_{0}^{x} e^{it^2y^2} y \, \mathrm{d}y \biggr] \, \mathrm{d}x \biggr). \end{align*}

To this end, we define the function $\varphi(\xi)$ by

$$ \varphi(\xi) = \int_{0}^{1} e^{i\xi s} \, \mathrm{d}s = \frac{e^{i\xi} - 1}{i\xi}. $$

Then $|\varphi(\xi)| \leq \int_{0}^{1} |e^{i\xi s}| \, \mathrm{d}s = 1 $ and $\varphi(\xi) \to 0$ as $|\xi| \to \infty$. Moreover, by substituting $y = \sqrt{s}$ for the first integral and $y = x\sqrt{s}$ for the second integral, OP's integral becomes

\begin{align*} I(t) &= e^{it^2} \biggl( \frac{\varphi(t^2)}{4} + \int_{0}^{1} \frac{1+3x^2}{(1+x^2)^2} \varphi(x^2 t^2) \, \mathrm{d}x \biggr). \end{align*}

If OP is aware of Dominated Convergence Theorem, the above equality immediately gives the desired answer by switching the order of $\lim_{t\to\infty}$ and $\int_{0}^{1} \ldots \, \mathrm{d}x$. Otherwise, we can still resolve the problem by elementary means.

Fix $\eta \in (0, 1)$ and note that

\begin{align*} |I(t)| &\leq \frac{|\varphi(t^2)|}{4} + \int_{0}^{1} \frac{1+3x^2}{(1+x^2)^2} |\varphi(x^2 t^2)| \, \mathrm{d}x \\ &\leq \frac{|\varphi(t^2)|}{4} + \int_{0}^{\eta} \frac{1+3x^2}{(1+x^2)^2} \, \mathrm{d}x + \Bigl( \sup_{s \geq \eta^2 t^2}|\varphi(s)| \Bigr) \int_{\eta}^{1} \frac{1+3x^2}{(1+x^2)^2} \, \mathrm{d}x. \end{align*}

Taking $\limsup$ as $t \to \infty$, we get

\begin{align*} \limsup_{t\to\infty} |I(t)| &\leq \int_{0}^{\eta} \frac{1+3x^2}{(1+x^2)^2} \, \mathrm{d}x. \end{align*}

Since the LHS is independent of $\eta$, letting $\eta \to 0^+$ yields $\displaystyle \limsup_{t\to\infty} |I(t)| = 0$ and we are done.

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Old Answer. The limit in the title and the body are different, so I will assume the one in the body is correct.

Substituting $x = y^2$,

$$ \int_{0}^{1} \frac{e^{it^2(1+y^2)}}{1+y^2} \, \mathrm{d}y = e^{it^2} \int_{0}^{1} \frac{e^{it^2 x}}{2\sqrt{x}(1+x)} \, \mathrm{d}x. $$

By the Riemann–Lebesgue lemma, the integral converges to $0$:

$$ \lim_{t \to \infty} \int_{0}^{1} \frac{e^{it^2 x}}{2\sqrt{x}(1+x)} \, \mathrm{d}x = 0. $$

Since the prefactor $e^{it^2}$ is bounded, this means that the entire expression converges to $0$.