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Differentiating Integrals

  • #1

Homework Statement


Given: f is integrable on [a,b] and has a jump discontinuity at c in (a,b) (meaning that both one-sided limits exist as x approaches c from the right and from the left but that Lim(x->c-)[f] is not equal to Lim(x->c+)[f]).

Show that F(x) = Integrate[f,{a,x}] is not differentiable at x = c (That is, F = the definite integral of f from a to x).



Homework Equations


N/A


The Attempt at a Solution



So we know that F is continuous by FTC - the only way that I can think of checking that F is not differentiable is... by showing that the derivative doesn't exist (using the limit). However, the limit reduces to a form that's rather unusable (it becomes Lim(x->c)[Integrate[f,{c,x}]/(x-c)], and I don't know how to deal with the Integral in the numerator).

Instead, I've been looking at those one-sided limits (which we know exist). If we knew that f was continuous apart from at this jump discontinuity, this problem would be simple. That fact is not given, however. My thinking is that, since the one-sided limits exist, F must be "one-sided continuous" kind of - that is, for the right hand limit, given epsilon > 0, there exists delta > 0 such that 0<x-c<delta implies |f(x)-L|<epsilon by definition (if L = the limit). Therefore, f is continuous slightly to the right of c. However, it's only continuous for an arbitrarily small segment (which changes with epsilon) - therefore, I can't really use this to make F'(x) = f(x) when 0<x-c<delta, can I?

Any help is appreciated.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
85
0
here is a way to think about it. a jump discontinuity will always look like a jump in the graph of f, right? so the integral up to c from the left will be an area. but past f this area is also going to make a jump right? therefore F is discontinuous so it is not differentiable.
 
  • #3
Thanks for you quick response!

Although this does provide the solution intuitively, I'm supposed to come up with a rigorous proof of this fact. Additionally, F actually is continuous at c (according to the fundamental theorem of calculus).

But, I agree, the area under the curve before c will be drastically different from the area under the curve after c. However, I'm not sure how I can make the difference between
Integral[f,{c,x}] when x is greater than c and Integral[f,{c,x}] when x is less than c explicit (that is, I'm not sure how to show that the limit as x->c will be different for the two integrals).
 
  • #4
Okay, so here's a solution that I just came up with (but I'm a bit iffy about using the limit to establish continuity near a point):

Lim(x->c+)[f] = L
=> Given epsilon > 0, there exists delta > 0 such that whenever 0<x-c<delta we have |f(x) - L| < epsilon/2.

Consider an arbitrary d such that 0<d-c< delta

Then:
|f(x) - f(d)|
= |f(x) - L + L - f(d)|
≤ |f(x) - L| + |f(d) - L|
< epsilon/2 + epsilon/2 = epsilon (assuming 0<x-c<delta)

Therefore, f is continuous at d and, by FTC, F'(d) = f(d).

Since the choice of d was arbitrary, F'(x) = f(x) for all x that satisfy 0<x-c<delta.

And therefore, Lim(x->c+)[F'(x)] = L.

Same idea for the left hand limit.

Any good? I feel like the definition of delta doesn't actually allow me to conclude continuity (though I could make a delta* = d-c or something similar, but then I feel like I have too many deltas depending on each other).
 
  • #5
85
0
hey, i see what you mean about using the FTC, but be careful the second version of the fundamental theorem is only applicable is f has an anti-derivative, otherwise we should assume it's continuous. here is a proof that does not require epsilon-del. Suppose that F is differentiable at x, then by the fundamental theorem F'(x)=f(x). thus F' is discontinuous and has a jump discontinuity at c. in other words lim x-->c F'(x) is different from either side; it doesn't exist. so F' is not continuous at c. so F is not differentiable at c. i may read over your proof later.
 
  • #6
Okay, so that proof doesn't work.

Here's why: when I'm given epsilon, I haven't actually chosen a value at which the function should be continuous. That is, I define the area that is continuous based on the choice of epsilon (and, therefore, delta). For this reason, the area that I'm proving "continuous" is only continuous for the one choice of epsilon - it's not always true, for any epsilon greater than zero.

However, I might have shown that, for all epsilon, there is an area around c where f is continuous. Which might be enough to show that the limit of F' is the same as the limit of f. I'm not sure...
 
  • #7
Oh, good call. I didn't even think of trying anything by contradiction. I think that what you've written works perfectly!

(My previous post was in regard to my proof!)

Thanks!!!
 
  • #8
85
0
wait wait wiat that could be wrong
 
  • #9
Hmmm... the only thing that bothers me about it is the assumption that f is continuous (except at c). Otherwise, it's good as far as I can tell...
 
  • #10
85
0
yeah i think you're right, since the limit exists from the right f is continuous near c, but i think that is essential.
 

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