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&#039;}^1{f(x)dx} &lt; \int_{h+\delta&#039;}^1{\frac{h} {h^2+x^2}}f(x)dx &lt; \frac{h} {h^2+(h+\delta&#039;)^2}\int_{h+\delta&#039;}^1{f(x)dx} (the maximum and minimum values of the fraction occur at h+\delta&#039; and 1 respectively). Therefore, since the integral is a number, <br /> \lim_{h\rightarrow 0^+} {\int_{h+\delta&#039;}^1{\frac{h} {h^2+x^2}}f(x)dx} = 0
and this is true for any \delta&#039;&gt;0.


Finally consider \int_{h}^{h+\delta&#039;}{\frac{h} {h^2+x^2}}f(x)dx. Add the requirement that h+\delta&#039; &lt; \delta so that <br /> \frac{h} {h^2+1}\int_{h}^{h+\delta&#039;}{f(x)dx} &lt; \int_{h}^{h+\delta&#039;}{\frac{h} {h^2+x^2}}f(x)dx &lt; (f(0)+\epsilon) \int_{h}^{h+\delta&#039;} {\frac{h} {2h^2}dx} = \frac{\delta&#039;}{2h}(f(0)+\epsilon) &lt; \frac{h}{2}(f(0)-\epsilon) if we also require that \delta&#039;&lt;h^2. Since we can make this smaller than any number by choosing small enough h and \delta&#039; with the given requirements, this shows that \lim_{h\rightarrow 0^+} {\int_{h}^{h+\delta&#039;}{\frac{h} {h^2+x^2}}f(x)dx} = 0 for any small enough \delta&#039;.


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&#039;}^1{\frac{h} {h^2+x^2}}f(x)dx} + 2\lim_{h\rightarrow 0^+} {\int_{h}^{h+\delta&#039;}{\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.