Proving a Double Integral Identity: Can Anyone Help with This Homework Problem?

[ggumdol]

Homework Statement

Can anybody prove the following double integral identity? How?:

\int_{0}^{1} s(1-s) f(sx) ds = \int_{0}^{1} s^2 \int_{0}^{1} t f(tsx) dt ds

Here f(x) is an arbitrary Riemann-integrable function.

Thanks in advance.

Homework Equations

I've found the following but it does not seem to be applicable to my problem.

http://onlinelibrary.wiley.com/doi/10.1002/9783527618132.app3/pdf

The Attempt at a Solution

I tried Coordinate Transformation and Integration by Parts. I've working on this problem for a couple of days. Please help me. I guess this identity could be proved simply by a coordinate transformation.

For those who may wonder if this identity is really correct, please run the following Maple code, then you will get 0, implying that the above identity is correct.

f(x) = x^2 - e^{x} * cos(x) + e^{x^2}+x^2 (cos (x))^2 + x^3 + 7 x^{5/4}

> f(x):=x^(2)-exp(x)*cos(x)+exp(x^(2))+x^(2)*(cos(x))^(2)+x^(3)+7*x^(5/(4));
> int(s*(1-s)*(eval(f(x), x = s*x)), s = 0 .. 1)-(int(s^2*(int(t*(eval(f(x), x = t*s*x)), t = 0 .. 1)), s = 0 .. 1));

 

[Dick]

Move the s^2 inside the inner integral. So your integral is s^2*t*f(stx)dt. Write that as (st)*f((st)*x)*d(st). Suggests you want a new variable of integration u=st, right? Ok, so now you've got a double integral du*ds. Mind the change in limits. Now change the order of integration.

 

[stevenb]

Homework Statement

Can anybody prove the following double integral identity? How?:

\int_{0}^{1} s(1-s) f(sx) ds = \int_{0}^{1} s^2 \int_{0}^{1} t f(tsx) dt ds

Here f(x) is an arbitrary Riemann-integrable function.

Thanks in advance.

Homework Equations

I've found the following but it does not seem to be applicable to my problem.

http://onlinelibrary.wiley.com/doi/10.1002/9783527618132.app3/pdf

The Attempt at a Solution

I tried Coordinate Transformation and Integration by Parts. I've working on this problem for a couple of days. Please help me. I guess this identity could be proved simply by a coordinate transformation.

For those who may wonder if this identity is really correct, please run the following Maple code, then you will get 0, implying that the above identity is correct.

f(x) = x^2 - e^{x} * cos(x) + e^{x^2}+x^2 (cos (x))^2 + x^3 + 7 x^{5/4}

> f(x):=x^(2)-exp(x)*cos(x)+exp(x^(2))+x^(2)*(cos(x))^(2)+x^(3)+7*x^(5/(4));
> int(s*(1-s)*(eval(f(x), x = s*x)), s = 0 .. 1)-(int(s^2*(int(t*(eval(f(x), x = t*s*x)), t = 0 .. 1)), s = 0 .. 1));

Of course, Dick always provides the elegant method.

I took a more pedestrian attempt, using a Taylor series representation of the function. I did this just to convince myself that your identity wasn't false, but I figured why not mention it.

If you can write f(y) as a Taylor series expansion, it is straightforward to prove.

You can use induction to simplify this by taking a general term f(y)=C*y^n (where C is a constant) and running it through the factory.

It's interesting that on one side you get something proportional to 1/(n+2)-1/(n+3) while on the other side you get something proportional to 1/((n+2)*(n+3)). Of course, it's not hard to show that these are equal.

 

[Dick]

Of course, Dick always provides the elegant method.

I took a more pedestrian attempt, using a Taylor series representation of the function. I did this just to convince myself that your identity wasn't false, but I figured why not mention it.

If you can write f(y) as a Taylor series expansion, it is straightforward to prove.

You can use induction to simplify this by taking a general term f(y)=C*y^n (where C is a constant) and running it through the factory.

It's interesting that on one side you get something proportional to 1/(n+2)-1/(n+3) while on the other side you get something proportional to 1/((n+2)*(n+3)). Of course, it's not hard to show that these are equal.

That's a valiant approach. Bravo! So you've shown directly it works for all functions of the form (stx)^n. I'll rate that approach one up from throwing a bizarre function at Maple to convince yourself it must be true.

 

[ggumdol]

Of course, Dick always provides the elegant method.

I took a more pedestrian attempt, using a Taylor series representation of the function. I did this just to convince myself that your identity wasn't false, but I figured why not mention it.

Thanks. I know that your (Tayler series expansion) method can be used to prove the identity but it may look slightly less elegant.

 

[ggumdol]

Homework Statement

Can anybody prove the following double integral identity? How?:

\int_{0}^{1} s(1-s) f(sx) ds = \int_{0}^{1} s^2 \int_{0}^{1} t f(tsx) dt ds

Here f(x) is an arbitrary Riemann-integrable function.

I got a complete solution due to hints from both of you.

\int_{0}^{1} s^2 \int_{0}^{1} t f(tsx) dt ds = \int_{0}^{1} \int_{0}^{1} s t f(tsx) s dt ds = \int_{0}^{1} \int_{0}^{s} u f(ux) du ds
where I plugged the change of variable u=st.

Now I change the area over which the double integration is done.
0 \leq u \leq s, ~ u \leq s \leq 1 ~ \rightarrow ~ 0 \leq u \leq 1 , ~ u \leq s \leq 1

Then we finally have
\int_{0}^{1} \int_{0}^{s} u f(ux) du ds = \int_{0}^{1} \int_{u}^{1} u f(ux) ds du = \int_{0}^{1} (1-u) u f(ux) du = \int_{0}^{1} (1-s) s f(sx) ds .
This simple identity has haunted me for a couple of days!

 

[Dick]

Thanks. I know that your (Tayler series expansion) method can be used to prove the identity but it may look slightly less elegant.

It doesn't PROVE the identity. As stevenb knows, not all functions can be expanded in a Taylor series. 'Riemann integrable' is way too weak for that. It's more of a plausibility argument like picking a strange function and testing it.

 

[Dick]

I got a complete solution due to hints from both of you.

\int_{0}^{1} s^2 \int_{0}^{1} t f(tsx) dt ds = \int_{0}^{1} \int_{0}^{1} s t f(tsx) s dt ds = \int_{0}^{1} \int_{0}^{s} u f(ux) du ds
where I plugged the change of variable u=st.

Now I change the area over which the double integration is done.
0 \leq u \leq s, ~ u \leq s \leq 1 ~ \rightarrow ~ 0 \leq u \leq 1 , ~ u \leq s \leq 1
Then we finally have
\int_{0}^{1} \int_{0}^{s} u f(ux) du ds = \int_{0}^{1} \int_{u}^{1} u f(ux) ds du = \int_{0}^{1} (1-u) u f(ux) ds du

Yep. That's it. You've got an extra ds at the end, but I know you are still editing.

 

[ggumdol]

Yep. That's it. You've got an extra ds at the end, but I know you are still editing.

Thanks Dick and stevenb. I like this forum.

 

[Dick]

Thanks Dick and stevenb. I like this forum.

Glad you do. I do too. If you want you can try and help other people with their problems. You're pretty sharp.