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Given a definite integral $$I=\int_a^b f(x)\text{d}x$$ for some function $f$, we can make any number of substitutions $x=u(x)$ which produce symbolically new integrals, all of which evaluate to the same value.

My main question is this: is it possible to have two definite integrals both of which give the same value yet each of which can not be transformed (i.e. made to look the same symbolically) into the other by substitution?

I get the impression the answer to my question is YES, but I'm not sure. Is there any information or theory out there that deals with this? How can we know when the transformation can not be done (if indeed it cannot be done)?

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What are the rules of the game here? That is, what types of algebraic manipulations do you want to consider? –  orlandpm Dec 12 '12 at 8:41
    
@orlandpm I'm just thinking of any algebraic substitution for the variable of integration. –  pbs Dec 12 '12 at 8:45
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Ok here's one that definitely provides a counterexample. Take $\rho$ to be a smooth function that vanishes in some interval $[0,a)$, with $0< a< 1$, and such that $\int_{0}^1\rho(x)\,dx = 1$. Now assume that we can find a substitution $u(x)$ such that $\rho(u(x)) \,du(x) = dx$ and $u(0) = 0$, $u(1) = 1$. This means that $\rho(u(x))u'(x)$ is identically $1$. But $\rho(u(x))$ must vanish in a neighborhood of $0$, so $\rho(u(x))u'(x)$ must vanish there too; this is a contradiction. –  Nick Strehlke Dec 12 '12 at 10:34
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Given any two definite integrals, you can multiply by constants to produce integrals that are equal. –  David Mitra Dec 12 '12 at 11:56
    
@DavidMitra +1, although I wanted to consider "non-trivial" substitutions which exclude multiplication by constants. –  pbs Dec 12 '12 at 12:00

3 Answers 3

The two definite integrals $$\int_0^\infty \sin(x^2)dx=\frac{\sqrt{2\pi}}{4}$$ and $$\int_0^\infty \cos(x^2)dx=\frac{\sqrt{2\pi}}{4}$$ are equal, but I know of no substitution which shows they are.

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I don't see immediately why there shouldn't exist such a substitution. Do you have a proof? –  Nick Strehlke Dec 12 '12 at 9:28
    
I don't have a proof. I did try to show these were equal using not only substitutions, but anything I could think of to work on the difference of the two integrals, without using contour integration. Nothing I tried like this showed they were equal. –  coffeemath Dec 12 '12 at 10:40

Let's start with the good-old unit semi-circle: $$I_1=\int_{-1}^{1} \sqrt{1 - x^2}\text{d}x = \frac{\pi}{2}$$

Now let's add a fancy triangle: $$I_2=\int_{0}^{1} \pi{x}\text{d}x = \frac{\pi}{2}$$

And, finally, a constant for good measure: $$I_3=\int_{0}^{1} \frac{\pi}{2}\text{d}x = \frac{\pi}{2}$$

Hey... what did you expect? It's 1:00am.

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Not bad for 1:00am. It's 8:50am here! –  pbs Dec 12 '12 at 8:50
    
We can convert $I_2$ to $I_3$ by the substitution $x = u^{1/2}$ in $I_2$. I don't think $I_1$ can be converted to $I_3$, but (unless I'm being dumb) this still requires proof. –  Nick Strehlke Dec 12 '12 at 9:56
    
You're, of course, absolutely right. I think what I meant to say was that you cannot convert $I_1$ to either $I_2$ or $I_3$ and it just came out wrong. As I said, 1:00am. That's my excuse and I'm sticking to it! –  Nik Bougalis Dec 12 '12 at 16:49

Note that, $$\int_0^{\pi /2}\cos x dx=1=\int_0^{\pi/2}(2/\pi)dx$$ Now for any substitution $x=u(x)$ we have $$\int_0^{\pi /2}\cos x dx=\int_0^{\pi/2}\cos (u(x))u^\prime(x)dx$$ So we should have, $2/\pi=\cos (u(x))u^\prime (x)$ which can not be true, because the LHS is constant but the RHS is not.

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This doesn't completely satisfy me. Why is it clear that we can't solve $u' = 2/(\pi \cos{u})$ on $(0,\pi/2)$, subject to $u(0) = 0$ and $u(\pi/2) = \pi /2$? Obviously we'd have to let $u'$ be infinite at $1$, but that might not be out of bounds. –  Nick Strehlke Dec 12 '12 at 9:27
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On second thought, we can solve $u' = 2/(\pi \cos{u})$ pretty easily. Take $u = \arcsin(2x/\pi)$. Then $u(0) = 0$ and $u(\pi/2) = \pi/2$, and also, since $\sin\circ u(x) = 2x/\pi$, the chain rule shows that $u' = 2/(\pi \cos{u})$. –  Nick Strehlke Dec 12 '12 at 9:52

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