I've been learning calculus II lately, and I was wondering (actually my teacher told me to find out), what is the proof of integration by substitution in the improper integral? I mean: the function $\phi(x)$ is continuous on the interval [a, b] and differentiable continuously on (a, b] with $\phi'(x)$ unlimited near the point a. The function $f(t)$ is continuous on the interval containing all values of $\phi(x)$. Three things that at this moment I don't know how to do:
- How to prove that: $ \int_{a}^{b}f(\phi(x))\phi'(x)\,dx$ is convergent.
- and how to prove that: $\int_{a}^{b}f(\phi(x))\phi'(x)\,dx = \int_{\phi(a)}^{\phi(b)}f(t) dt \\$
- How the prove will change if b = $\infty$?
Can anyone give me dome hints or directions on how to do this? Thank you in advance!
As you noted, the range of $ \phi $ is an interval (by the Intermediate-Value Theorem), and $ f $ is continuous on that interval, so it has an antiderivative $ F $ there. Using this, you can get an antiderivative $ \int f ( \phi ( x ) ) \phi ' ( x ) \, \mathrm d x $; this will be an expression defined for $ x \in [ a , b ] $, but it's only an antiderivative for $ x \in ( a , b ] $, since $ \phi ' ( x ) $ is only defined there. But you can use this to find $ \int _ { x = x _ 0 } ^ b f ( \phi ( x ) ) \phi ' ( x ) \, \mathrm d x $ for any $ x _ 0 $ in $ ( a , b ] $. And the improper integral that you want is defined as the limit of this as $ x _ 0 \to a ^ + $. So you just have to check that your expression for $ \int _ { x = x _ 0 } ^ b f ( \phi ( x ) ) \phi ' ( x ) \, \mathrm d x $ is continuous at $ x _ 0 = a $ to answer (1), and that its value at $ x _ 0 = a $ equals $ \int _ { t = \phi ( a ) } ^ { \phi ( b ) } f ( t ) \, \mathrm d t $ to answer (2).
For (3), if $ b = \infty $, then you have an intergral that's improper in two directions, so it might be simpler to look at $ a = - \infty $ instead (or redo (1) and (2) with $ \phi ' $ defined on $ [ a , b ) $ instead of on $ ( a , b ] $). (But the two improper directions are independent, so the way you wrote it still makes sense.) Now, it doesn't make sense to say that $ \phi $ is continuous on $ [ a , b ] $ (unless you're seriously working in the extended real-number system in your course), so I think that the best that you could do is to say that $ \phi $ is continuous on $ [ a , b ) $ and the limit $ L = \lim _ { x \to \infty } \phi ( x ) $ exists. So then you'd want $ f $ to be continuous not only on the range of $ \phi $ but also at $ L $. (Notice that $ \operatorname { ran } \phi \cup \{ L \} $ is still an interval.) Then you can continue as before, using $ L $ instead of $ \phi ( b ) $.
You could also consider a more general version of (3) where $ f $ is not defined at $ L $; for that matter, you could consider a more general version of (1) and (2) where $ f $ is not defined at $ \phi ( a ) $. Then your answer $ \int _ { x = \phi ( a ) } ^ { \phi ( b ) } f ( t ) \, \mathrm d t $ turns out to be an improper integral as well. It gets more complicated if you want to consider what happens when $ L $ or $ \phi ( a ) $ doesn't exist (or when $ \phi $ is not continuous at $ a $).
Hopefully this gives you some idea of how to go about this, without giving away the answer completely!