Proof for the limit of a definite integral where the integrand varies

[Boorglar]

Homework Statement

This problem was taken from Spivak's Calculus 3rd Edition, Problem 19-25 (d).

Let $f$ be integrable on $[-1 , 1]$ and continuous at $0$. Show that $$\lim_{h\rightarrow 0^+} {\int_{-1}^{1}{\frac{h} {h^2+x^2}}f(x)dx = {\pi}f(0).}$$

Homework Equations

I already proved from part (c) that $$\lim_{h\rightarrow 0^+} {\int_{-1}^{1}{\frac{h} {h^2+x^2}}dx = {\pi}}$$
which is easy with arctan.

The Attempt at a Solution

I found a way to prove it but I am not 100% sure that it is rigorous enough (and the proof given in the answers is different).

First consider $\int_{-h}^{h}{\frac{h} {h^2+x^2}}f(x)dx.$ Since $f$ is continuous at $0$, then for some $\delta > 0,$ $f(0) - \epsilon < f(x) < f(0) + \epsilon$ for all x in $[-\delta , \delta].$ Choose $0<h<\delta$. Then $$(f(0)-\epsilon) \int_{-h}^{h}{\frac{h} {h^2+x^2}}dx < \int_{-h}^{h}{\frac{h} {h^2+x^2}}f(x)dx < (f(0)+\epsilon) \int_{-h}^{h}{\frac{h} {h^2+x^2}}dx$$ $${\pi}(f(0)-\epsilon) ≤ \lim_{h\rightarrow 0^+} {\int_{-h}^{h}{\frac{h} {h^2+x^2}}f(x)dx} ≤ \pi(f(0)+\epsilon)$$ taking the limits by part (c) and this is true for any $\epsilon > 0$ so $$\lim_{h\rightarrow 0^+} {\int_{-h}^{h}{\frac{h} {h^2+x^2}}f(x)dx} = {\pi}f(0)$$

Now consider $\int_{h+\delta'}^1{\frac{h} {h^2+x^2}}f(x)dx$ for any $\delta' > 0.$
$$\frac{h} {h^2+1}\int_{h+\delta'}^1{f(x)dx} < \int_{h+\delta'}^1{\frac{h} {h^2+x^2}}f(x)dx < \frac{h} {h^2+(h+\delta')^2}\int_{h+\delta'}^1{f(x)dx}$$ (the maximum and minimum values of the fraction occur at $h+\delta'$ and $1$ respectively). Therefore, since the integral is a number, $$\lim_{h\rightarrow 0^+} {\int_{h+\delta'}^1{\frac{h} {h^2+x^2}}f(x)dx} = 0$$
and this is true for any $\delta'>0.$


Finally consider $\int_{h}^{h+\delta'}{\frac{h} {h^2+x^2}}f(x)dx.$ Add the requirement that $h+\delta' < \delta$ so that $$\frac{h} {h^2+1}\int_{h}^{h+\delta'}{f(x)dx} < \int_{h}^{h+\delta'}{\frac{h} {h^2+x^2}}f(x)dx < (f(0)+\epsilon) \int_{h}^{h+\delta'} {\frac{h} {2h^2}dx} = \frac{\delta'}{2h}(f(0)+\epsilon) < \frac{h}{2}(f(0)-\epsilon)$$ if we also require that $\delta'<h^2$. Since we can make this smaller than any number by choosing small enough $h$ and $\delta'$ with the given requirements, this shows that $$\lim_{h\rightarrow 0^+} {\int_{h}^{h+\delta'}{\frac{h} {h^2+x^2}}f(x)dx} = 0$$ for any small enough $\delta'$.


Combining all three results shows that $$\lim_{h\rightarrow 0^+} {\int_{-1}^{1}{\frac{h} {h^2+x^2}}f(x)dx} = \lim_{h\rightarrow 0^+} {\int_{-h}^{h}{\frac{h} {h^2+x^2}}f(x)dx} + 2\lim_{h\rightarrow 0^+} {\int_{h+\delta'}^1{\frac{h} {h^2+x^2}}f(x)dx} + 2\lim_{h\rightarrow 0^+} {\int_{h}^{h+\delta'}{\frac{h} {h^2+x^2}}f(x)dx} = {\pi}f(0) + 0 + 0 = {\pi}f(0)$$ (using symmetry).

QED


Ok that was long and maybe confusing, I'm sorry... but that's the proof I could find. Can you tell me if there are any gaps or incorrect assumptions in the proof? Thanks!

 

[tt2348]

I see that for $\lim_{h\rightarrow 0^+} {\int_{-1}^{1}{\frac{h} {h^2+x^2}}dx = {\pi}}$ , but $\int_{-h}^{h}{\frac{h} {h^2+x^2}}dx= {\pi}/2$ I believe? If you visualize $\lim_{h\rightarrow 0^+}$ as $\lim_{n\rightarrow \inf}$ with all the h=1/n, you can evaluate as a pointwise convergent function, that only has value at x=0 as n-> inf, and if you normalize ${\frac{h} {h^2+x^2}}$ over ℝ , you'll get a delta function, which gives useful properties in terms of integration. Nascent delta function comes to mind for this case

 

[Boorglar]

I see that for $\lim_{h\rightarrow 0^+} {\int_{-1}^{1}{\frac{h} {h^2+x^2}}dx = {\pi}}$ , but $\int_{-h}^{h}{\frac{h} {h^2+x^2}}dx= {\pi}/2$ I believe?

Oh you're right! Then I must've completely messed-up haha (although I still got the correct answer)

In this case this would mean that another ${\frac {\pi} {2}}f(0)$ must appear from somewhere... And yet intuitively the other integrals should converge to 0, shouldn't they?

 

[tt2348]

Replace your limits of integration with h^2, (or 1/n^2) ... i loaaaaaaaathe one sided limits, and prefer sequences of 1/n that will converge to 0+.
Remember youre dealing with a function that pointwise will converge to 0 for all non zero x, but go to infinity for x=0. if you redefine youre h/(h^2+x^2) as a delta function by normalizng the integral over R ( ie delta(X)=lim n-> inf n/(pi*(1+(nx)^2)), youll get integral delta(x)*f(x)=f(0), and multiplying the pi over from the normalized delta function gives pi*f(0)

 

[Boorglar]

hmmm I'm not really familiar with the delta function. Why does the delta function work with those kinds of limits? And also, since the book doesn't mention it yet, I suppose he expects a solution using an epsilon-delta argument.

 

[tt2348]

Delta function is 0 for all non zero x, but goes to infinity at x=0. It's interesting because when integrated over all R, it's 1, and pops up a lot in physics. What material are you covering in this chapter? Anything having to do with cauchys principal value theorem?

 

[Boorglar]

No it's actually a chapter on Indefinite Integration. But the problems tend to get hard and more off-topic after the first few. I haven't heard of the Cauchy principal value theorem
In the answers he uses an epsilon-delta argument but his proof was confusing to me so I tried finding a similar argument that would be more clear to me.

 

[tt2348]

Is there any way you could post the solution he gave? I could help explain what he is doing. If not, there's a whole complex analysis way of solving this also.