How can I solve this limit with a definite integral inside?

This is the problem:

$$\lim_{m \to 0}\left(\frac1m \int_{\sqrt\frac\pi 2}^{\sqrt\frac\pi 2+m}\sin(x^2)\,dx\right)$$

I don't have an idea how to solve it. I was thinking maybe it's an indeterminate form $$\infty \cdot 0$$ and we can do something with l'hopital's rule. But I don't know what to do with the integral. What is the trick here?

• Maybe it helps to define $F(m)= \int_{\sqrt\frac\pi 2}^{\sqrt\frac\pi 2+m}(sinx^2dx)$ and use the fundamental theorem of calculus. – Ilovemath Jan 15 at 21:12
• Are you familiar with the Lebesgue Differentiation Theorem? – DMcMor Jan 15 at 21:19
• One does not need the general Lebesgue differentiation theorem since the integrand is smooth. – user9464 Jan 15 at 21:28

Observe that we have both

$$\lim_{m\to0} m=0\;\;\;\text{ and}\;\;\;\lim_{m\to0}\int_{\sqrt\frac \pi2}^{\sqrt\frac \pi2+m}\sin x^2\,dx=0$$

so by L'Hospital's rule, with $$\;F(x)\;$$ being a primitive function of $$\;\sin x^2\;$$ :

$$\lim_{m\to0}\frac{\int_{\sqrt\frac \pi2}^{\sqrt\frac \pi2+m}\sin x^2\,dx}m=\lim_{m\to0}\left[F\left(\sqrt{\frac \pi2}+m\right)-F\left(\sqrt{\frac \pi2}\right)\right]'=$$

$$=\lim_{m\to0}F'\left(\sqrt{\frac \pi2}+m\right)=\lim_{m\to0}\sin\left(\sqrt{\frac \pi2}+m\right)^2=\sin\frac\pi2=1$$

In the above, the derivative is with respect to $$\;m\;$$ , of course.

• Can you please explain why $$\lim_{m\to0}\left[F\left(\sqrt{\frac \pi2}+m\right)-F\left(\sqrt{\frac \pi2}\right)\right]'=\lim_{m\to0}F'\left(\sqrt{\frac \pi2}+m\right)$$ ? – Quaei Jan 15 at 22:43
• The derivative with respect to $\;m\;$ on the left side equals the derivative in the right side. Observe that the second therm on the left side is a constant number and thus its derivative equals zero – DonAntonio Jan 15 at 22:48

The result follows from the fundamental theorem of calculus. The limit is equal to $$\frac{d}{dx}\int_a^{x}\sin(y^2)\,dy\bigg|_{x=a}=\sin(a^2)\tag{1}$$

If you use l'Hopital, then the limit is equal to $$\lim_{m\to 0}\frac{\dfrac{d}{dm}\int_a^{a+m}\sin(x^2)dx}{1}$$ and you end up with the same result in (1).

• FTC is the right approach. L'Hospital's is pretty roundabout here. – Paramanand Singh Jan 16 at 1:22
• @ParamanandSingh There is no "right" or "wrong" approach here. There is only correcto or incorrect, and in this case both FTC and L'Hospital are correct. Besides this, the very first line is in fact the definition of derivative at $\;\sqrt\frac\pi2\;$, and L'Hospital goes perfectly well with this. Now, if the requirements were not to use L'H then I'd agree. – DonAntonio Jan 16 at 8:56
• @DonAntonio: I did not mean to say that L'Hospital's is incorrect, but rather more work. The result is immediate by FTC. But with L'Hospital's you first apply FTC to find derivatives and then take limit. Also I had given an upvote already as I agreed with the overall answer. I think we are on same page here. – Paramanand Singh Jan 16 at 9:31
• @ParamanandSingh You say "the result is immediate by FTC"...but this is not completely accurate: first you must understand and deduce that the wanted limit is the definition of the derivative of the given integral at $\;\sqrt\frac\pi2\;$, otherwise you get nothing.. So there is also more work here than merely using FTC... – DonAntonio Jan 16 at 9:37
• @DonAntonio: FTC says that if $F(x) =\int_a^x f(t) \, dt$ then $F'(x) =f(x)$ at points of continuity of $f$. Now by definition of derivative this means that $\lim_{h\to 0}\frac{1}{h}\int_c^{c+h}f(t)\,dt=f(c)$. So I would consider it an immediate consequence of FTC. – Paramanand Singh Jan 16 at 9:41

Another way to do it.

Let $$x=t+\sqrt{\frac{\pi }{2}}$$ to make $$\int_{\sqrt\frac\pi 2}^{\sqrt\frac\pi 2+m}\sin(x^2)\,dx=\int_0^m \cos \left(\sqrt{2 \pi } t+t^2\right)\,dt$$ Now, By Taylor and binomial expansion $$\cos \left(\sqrt{2 \pi } t+t^2\right)=1-\pi t^2-\sqrt{2 \pi } t^3+O\left(t^4\right)$$ $$\int \cos \left(\sqrt{2 \pi } t+t^2\right)\,dt=t-\frac{\pi t^3}{3}-\frac{1}{2} \sqrt{\frac{\pi }{2}} t^4+O\left(t^5\right)$$ $$\int_0^m \cos \left(\sqrt{2 \pi } t+t^2\right)\,dt\sim m-\frac{\pi m^3}{3}+O\left(m^4\right)$$ $$\frac 1m\int_0^m \cos \left(\sqrt{2 \pi } t+t^2\right)\,dt\sim 1-\frac{\pi m^2}{3}+O\left(m^3\right)$$

Just for the fun, computing the exact value of the integral for $$m=0.1$$ (far away from $$0$$), the expression would be $$0.9889$$ while the approximation gives $$1-\frac{\pi }{300}=0.9895$$.