Thread: Integral of even functions

Please help me with this proof.

f(x) is even on [-a, a] and integrable on [0, a]; i.e. there exists int(0,a;f(x)dx).

Prove that int(-a,a;f(x)dx) = 2*int(0,a;f(x)dx).


Thanks for any help,

Re:

In the above, did you assume that Int[-a,0]f(x)dx exsits?
I think that we need to prove that.

Since I showed that Int[-a,0]f(x)dx=Int[0,a]f(-x)dx, where f is integrable on [0,a], I did prove it, don't I?

Originally Posted by alinailiescu

Since I showed that Int[-a,0]f(x)dx=Int[0,a]f(-x)dx, where f is integrable on [0,a], I did prove it, don't I?

No, I don't think so. Look at what you did.
You assumed Int[-a,0]f(x)dx existed and did a change if variable on it.
All we are given is that Int[0,a]f(x)dx exists.

Thanks alinailiescu and Plato for your answers.

But in your development, alinailiescu, when you changed from variable u to x, something hit me:

int(0, a);f(-u)du = int(0, a);f(-x)dx

In my opinion (I'm a newbie) you should substitute -u by x (because you set u = -x) and du by -dx. If not which could be the explanation ?

Thanks a lot for your attention!

You cannot use change of variable. I believe you need the facts that: u=-x is differenciable on (-a,a) and u' is continous on (-a,a) which is true. But you also need to know that f is continous on [-a,a]. The problem does not state that, it just states that f is integrable not necessarily continous.

I will try to do this problem without using the substitution theorem because it is faulty here.

Originally Posted by CalculusFan

Please help me with this proof.

f(x) is even on [-a, a] and integrable on [0, a]; i.e. there exists int(0,a;f(x)dx).

Prove that int(-a,a;f(x)dx) = 2*int(0,a;f(x)dx).


Thanks for any help,

Consider any partition P of [-a,0], then there is a corresponding partition
P' = -P of [0,a], and vice versa. Also as f is even the upper and lower
Reimann sums for the partitions P and P' and f are equal, hence if one
limit exists as the length of the longest interval in such partitions goes to
zero so does the other, and hence both integrals exist and are equal.

(You may need to tightent this up a bit)

RonL

Originally Posted by CaptainBlank

Consider any partition P of [-a,0], then there is a corresponding partition
P' = -P of [0,a], and vice versa. Also as f is even the upper and lower
Reimann sums for the partitions P and P' and f are equal, hence if one
limit exists as the length of the longest interval in such partitions goes to
zero so does the other, and hence both integrals exist and are equal.

(You may need to tightent this up a bit)

RonL

I was thinking about this too. In fact, my Calculus book (if I remember correctly) uses this proof instead. I am just saying there is something wrong with substitution rule here, I can feel it in my bones.

Thank you all people!

All the reasoning you presented added to my understanding of this defying area of Math. But CapitainBlank's argument put an end to the discussion that I'm having with my mates: we all, finally, agree with the proof.

Even though it is true that the condition that f be continuous is found in most calculus text and even beginning analysis text, it is sufficient that f be integrable with additional conditions on u. Jerrold Marsden has a good discussion in his text on analysis.

This should be a warning to all: don’t assume too much.
Clearly RonL’s solution is what was looked for.
I for one mistakenly assumed the question was aimed at a different level.

Originally Posted by Plato

This should be a warning to all: don’t assume too much.

I am a formalist, that does not apply to me.
---
Anyway, here is what CaptainBlank was saying.
With my commentaries.

Note, I will use the Darboux Integral.
Which can also be found here.

Theorem: Let f be integrable on [0,a] (for a>0) and define g(x) = f(-x) for all x in [0,a]. Then g(x) is integrable on [-a,0] and furthermore its integral is equal to the integral of f on [0,a].

Proof: The fact the g is bounded is easily seen. Now follow the attachment.

I frankly do not understand your point!
The Darboux Integral is equivalent to the Riemann Integral.
Remembering that you are using Ken Ross’s book, look at his chapter of the Integral. Ross says that he has entitled the section ‘Riemann Integral’ ever though he will use the Darboux approach because it is equivalent to that of Riemann.

Originally Posted by Plato

I frankly do not understand your point!
The Darboux Integral is equivalent to the Riemann Integral.
.

I was just doing this out in full detail. For others to see the proof.

This well-known result can be found in every college calculus book.