# Conditions that allow Integration by Substitution

EDITED

If asked to integrate$$\int_a^b{f(x)}dx$$ what kind of functions can I exactly substitute $x$ with?

I believe that if let's say $x=g(x)$ is integrable and $f(x)$ is continuous then integrability is retained. Does it retain equality of integrals after substitutions?

Is it different if we change $$\int_a^bf(x)\, \mathrm{d}x \text{ to} \int^b_af\, \mathrm{d}\alpha$$ I mean are conditions different for riemann and stieltjes integrals?

Or, precisely, what are minimum conditions necessary to be allowed to substitute in integrals? I believe bicontinuous bijections (??) are sufficient.

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By minimum, you mean the necessary and sufficient condition of $\phi$ such that for each $f$ is (Riemann/Lebesgue)-integrable on $[a,b]$, we have $\int_a^b f(\phi(x))\phi^\prime(x)dx=\int_{\phi(a)}^{\phi(b)}f(x)dx$. It's quite hard. I know that it's true when $\phi$ is monotone. – Frank Science May 10 '13 at 17:40
thanks. got answer after months. but great. – user45099 May 10 '13 at 20:35

As you use the Chain-rule for this one you function must necessary be continuous differentiable. But if you substitue don't forget to change you bounds (correct me if it is the wrong word). As a formel $$\int_a^b f(\varphi(t)) \cdot \varphi'(t) \, \mathrm{d}t = \int_{\varphi(a)}^{\varphi(b)} f(x)\, \mathrm{d}x$$
what you did is $\int^b_af(\phi(t))d(\phi(t))=\int^b_af(\phi(t))\phi'(t)dt.$ This very step assumes differentiability of $\phi$. My point is $\int^b_af(\phi(t))d(\phi(t))$ can be integrable even if the above identity does not hold or precisely even if, $\phi$ is not differentiable. Disregard by ignoring of bounds. I am too lazy to correct them. – user45099 Feb 20 '13 at 19:17
it is integrable if $f(\varphi(t))$ is a regulated function, but the values of the integral will be completly different – Dominic Michaelis Feb 20 '13 at 19:24
Ok, so if I got it correct, $\phi$ should be continuous and differentiable to allow the integrals to be equal. I know it is sufficient but I am looking for necessary conditions. That is why I am being stringent. – user45099 Feb 20 '13 at 19:31
$\phi$ needn't to be continuously differentiable. For example, if $\phi$ is monotone and differentiable, where $\phi^\prime$ is integrable, the theorem is also correct. @ήλιος – Frank Science May 10 '13 at 17:26