6
$\begingroup$

Edited: changed $\displaystyle\int_{a}^{b}f(g(t))g'(t) \, dt = \int_{g(a)}^{g(b)}f(x) \, dx$ TO $\displaystyle\int_{a}^{b}f(t) \, dt = \int_{f(a)}^{f(b)}u \, \frac{du}{f'(f^{-1}(u))}$

I am wondering about this idea in general, but for concreteness let's say I am attempting to numerically integrate the following:

$$ \int_{-\pi}^{\pi} K\left(\sqrt{1-\frac{1}{4}\left(\sqrt{3}\varepsilon+\cos(x)\right)^2}\right)dx=\int_{-\pi}^{\pi} K\left(\alpha\right)dx $$ Where $K$ is the elliptic integral of the first kind, $-\sqrt{3}<\varepsilon<\sqrt{3}$ is a parameter and for compactness I've defined $\alpha\equiv \sqrt{1-\frac{1}{4}\left(\sqrt{3}\varepsilon+\cos(x)\right)^2}$.

Let's say I want to make a change in variables before trying to crunch this numerically, for whatever reason. Let's also say I have settled on the substitution $u^3=\alpha-1$.

My question is how do I determine the bounds on the new integral? I know (thought I knew?) that in general I just solve $u^3=\alpha-1$ for $x=\pi$ and $x=-\pi$ to get my new bounds. In this case however, that results in both bounds being the same. No problem, I notice the integrand is symmetric in $x$ so I can do twice the integral from $0$ to $\pi$, which gives me different $u$-bounds. My question then is when are we allowed to apply: $$ \int_{a}^{b}f(t) \, dt = \int_{f(a)}^{f(b)}u \, \frac{du}{f'(f^{-1}(u))}\quad \text{ with } \ u=f(t) \to dt=\frac{du}{f'(f^{-1}(u))} $$ if it doesn't work when $u(a)=u(b)$? Furthermore, even going from $0$ to $\pi$ I run into the same problem when $\varepsilon=0$. I could again exploit symmetry to break up the integral, or scrap this idea entirely and try a different substitution, but I am looking for a more general method here or some insight into where this failure comes from. For example, if I let $|\varepsilon|\ll1$ my bounds in $u$ are different, but just barely. I would expect this to fail as well even though the bounds are different. It seems like there must be some need to take into account the shape of $g(t)$, perhaps $g'(t)\geq0$ or $g'(t)\leq0$ for $a<t<b$ or something like that? Any insight is appreciated, thanks in advance!

$\endgroup$
  • 1
    $\begingroup$ This is a good question. I thought I could answer it by just consulting the formal change of variables theorem but I came up empty. I hope it gets some more attention. $\endgroup$ – JonathanZ supports MonicaC Sep 25 '18 at 0:13
3
$\begingroup$

The change of variables result

$$\tag{*}\int_{a}^{b}f(g(t))g'(t) \, dt = \int_{g(a)}^{g(b)}f(x) \, dx$$

is sometimes proved in texts with the hypotheses that $f$ is continuous on $g([a,b])$ and $g$ has a continuous derivative $g'$ on $[a,b]$ which is never zero. In this case, $g$ is strictly increasing or decreasing and the issue that $g(a) = g(b)$ never arises.

However, the condition that $g'(x) \neq 0$ can be discarded and (*) is still valid. This holds even if $g(a) = g(b)$ where the integral must be zero.

For a proof, note that $t \mapsto f(g(t))g'(t)$ is continuous and we can define

$$F(x) = \int_{g(a)}^x f(t) \, dt,\\G(x) = \int_a^x f(g(t))g'(t) \, dt$$

Thus, $F'(x) = f(x)$, $G'(x) = f(g(x)) g'(x)$ and by the chain rule

$$\frac{d}{dx}F(g(x)) = f(g(x))g'(x) = G'(x)$$

It follows that $F(g(x)) - G(x)$ is a constant. Since $F(g(a)) = G(a) = 0$ the constant is zero and for all $x \in [a,b]$ we have

$$G(x) = \int_a^x f(g(t))g'(t) \, dt = \int_{g(a)}^{g(x)} f(t) \, dt = F(g(x))$$

This is true, in particular for $x = b$ where we recover (*) even if $g(a) = g(b)$.

Also there are even weaker conditions on $f$ and $g$ where this result is true.

Revised question

The question is how does "u-substitution" as a variant of the change-of-variables theorem break down when applied to $\int_a^b f(t) \, dt$ using the formal approach: $ u = f(t), \,\, du = f'(t) \, dt$.

In defining $u = f(t)$ and proceeding with the substitution we require that $f$ is injective (the inverse function $f^{-1}$ exists with range containing [a,b]).

We then can apply the usual formula for the derivative of the inverse function to $t = f^{-1}(u)$ and obtain

$$dt = \frac{d}{du}[f^{-1}(u)] \, du = \frac{1}{f'(f^{-1}(u))} \, du,$$

and

$$\int_a^b f(t) \, dt = \int_{f^{-1}(a)}^{f^{-1}(b)} \frac{u}{f'(f^{-1}(u))} \, du$$

This is is what I believe you intended to write. It is only valid if the derivative in the denominator on the RHS is never zero and

$$f([f^{-1}(a), f^{-1}(b)]) = [a,b].$$

$\endgroup$
  • 2
    $\begingroup$ You made a mistake so this is not a counterexample. In this case $g'(t) = -2 \cos t \sin t$. . $\endgroup$ – RRL Sep 25 '18 at 14:56
  • 2
    $\begingroup$ In fact $\int_0^{\pi }\cos^2 t [-2 \cos (t) \sin (t)] \, dt= \int_1^1 dt = 0$. $\endgroup$ – RRL Sep 25 '18 at 14:59
  • 1
    $\begingroup$ I think made a mistake in my question! Let's say I want to do a $u$-substitution, not necessarily $f(g(t))g'(t)dt$, like: $$\int_0^{\pi} \sin(x) \,\mathrm{d}x=2$$ If I now say $u=\sin(x)\to \mathrm{d}u=\cos(x)\mathrm{d}x=\sqrt{1-u^2}\mathrm{d}x$ and $x=0\to u=0$, $x=\pi\to u=0$ so that $$\int_0^{\pi} \sin(x) \,\mathrm{d}x=2 \to \int_0^{0}\frac{u \, \mathrm{d}u}{\sqrt{1-u^2}}=0$$ This is the $u$-substitution failure I was referring to, sorry my original question doesn't reflect that. I am new to stack exchange, should I post this as a new question? Thanks again! $\endgroup$ – bRost03 Sep 25 '18 at 15:09
  • 1
    $\begingroup$ I see you are correct when the integrand is of the form $f(g(t))g'(t) \, \mathrm{d}t$, which follows very simply from the fundamental theorem of calculus as you showed. $\endgroup$ – bRost03 Sep 25 '18 at 15:17
  • 1
    $\begingroup$ I have edited my question to reflect this, sorry for the confusion and thanks again for your help! $\endgroup$ – bRost03 Sep 25 '18 at 15:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.