Proving $-\frac{1}{a}<\int_a^b \sin(x^2) dx<\frac{1}{a}$

Question

I have encountered a question:

Prove $$-\frac{1}{a}<\int_a^b \sin(x^2) dx<\frac{1}{a}$$

There are plenty of solutions to $\int_0^{\infty} \sin(x^2) dx$ online, but there seems to be no solution to the boundary of $\int_a^b \sin(x^2) dx$.

Could anyone help me with this please? I tried to calculate the integral directly, but I cannot cancel out b and get a boundary only with $a$.

Applying inequality to the integrand $\sin(x^2)<x^2$ does not work either.

@Jack, the last inequality is not true. Taking absolute value is not going to work, you are not using the cancellation that arises from the oscillation of $\sin$. — Julian Mejia, Jun 11 '19 at 02:32
@Jack that gives an estimate of the integrand not the integral. If you want to bound the integral then you should also multiply the length of the interval where you are integrating. — Julian Mejia, Jun 11 '19 at 02:52
Hmm, I indeed only had the bound $\frac{b^2-a^2}{2a}$ instead of the one I claimed. — , Jun 11 '19 at 02:53
I think I managed to do it, as I mentioned earlier, we need to take advantage of the oscilation of $\sin$, for this we use integration by parts. — Julian Mejia, Jun 11 '19 at 03:25

score 12 · Answer 1 · edited Jun 11 '19 at 03:37

By change of variables $u=x^2$, we have $\displaystyle{\int_{a}^{b}\sin(x^2)\,{\rm d}x=\frac{1}{2}\int_{a^2}^{b^2}\frac{\sin(u)}{\sqrt{u}}\,{\rm d}u}$.

Now, by integration by parts we have $$\int_{a^2}^{b^2}\frac{\sin(u)}{\sqrt{u}}\,{\rm d}u=-\int_{a^2}^{b^2}\frac{1}{\sqrt{u}}{\rm d}(\cos u)=\frac{\cos(a^2)}{a}-\frac{\cos(b^2)}{b}-\frac{1}{2}\int_{a^2}^{b^2}\frac{\cos u}{u^{3/2}}{\rm d}u$$

Now, note that $$\left|\frac{\cos(a^2)}{a}\right|\leq \frac{1}{a},$$ $$\left|\frac{\cos(b^2)}{b}\right|\leq \frac{1}{b}$$ and $$\frac{1}{2}\left|\int_{a^2}^{b^2}\frac{\cos u}{u^{3/2}}{\rm d}u\right|< \frac{1}{2}\int_{a^2}^{b^2}\frac{1}{u^{3/2}}{\rm d}u=\frac{1}{a}-\frac{1}{b}.$$

Hence, $\displaystyle{\left|\int_{a^2}^{b^2}\frac{\sin(u)}{\sqrt{u}}{\rm d}u\right|<\frac{1}{a}+\frac{1}{b}+\left(\frac{1}{a}-\frac{1}{b}\right)=\frac{2}{a}}$. We conclude $$\left|\int_{a}^{b}\sin(x^2)\,{\rm d}x\right|=\frac{1}{2}\left|\int_{a^2}^{b^2}\frac{\sin(u)}{\sqrt{u}}{\rm d}u\right|<\frac{1}{a} $$

@Jack I am assuming what the OP mentioned $0<a<b$( so $0<a^2<b^2$). For short, the equality could only happen if $|\cos u|=1$ a.e. in $[a^2,b^2]$ but that cannot happen. — Julian Mejia, Jun 11 '19 at 13:52

Proving $-\frac{1}{a}<\int_a^b \sin(x^2) dx<\frac{1}{a}$

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