# Is this true about integrating composite functions?

Let's say that I'm integrating a composite function, say $f(g(x))$, that is in a form to which I can apply the substitution rule. Is it true to say that both $f$ and $g$ must be differentiable?

I understand that the substitution rule requires $g$ to be differentiable and that the substitution rule relies on the chain rule, and the chain rule requires both f and g to be differentiable.

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Suppose we want to do $\int_0^1 e^{-1/x}dx$. With $u=1/x$, this becomes $\int_1^\infty e^{-u}u^{-2}du$. The function $g(x)=1/x$ is nondifferentiable at $x=0$, but this isn't a problem. The problem here may be that you probably have a certain theorem in mind, and you want to know whether a certain assumption in the theorem is really necessary -- but you haven't stated the theorem. – Ben Crowell Feb 7 '12 at 5:44

if you talk about the Riemann-integral: if $f : [a,b] \to [c,d]$ R-integrable and $g : [c,d] \to \mathbb{R}$ continuous, than $g \circ f : [a,b] \to \mathbb R$ is R-integrable. Differentiable implies continuous, so if f,g are differentiable they are R-integrable

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Substitution rule relies on chain rule and the chain rule requires both f and g to be differentiable.

In showing $\displaystyle \int f(x)dx=\displaystyle \int f(g(t))g'(t)dt$,

All we require in substitution rule is that $f(x)$ be continuous so that its antiderivative exists-and is differentiable, say $F$ and $g(x)$ to have a continuous derivative, so the composition $F \circ g$ is differentiable, which we differentiate using chain rule.

So, by chain rule, $\displaystyle\frac{d}{dt}(F \circ g)(t)=f(g(t))g'(t)$. By fundamental theorem of calculus,

$\displaystyle \int_a^b f(g(t))g'(t)dt=(F \circ g)(b)-(F \circ g)(a)=\displaystyle \int_{g(a)}^{g(b)}f(x)dx$. Thus we have proved substitution rule.

Since we don't need to differentiate $f \circ g$ to prove substitution rule, we don't require $f$ to be differentiable.

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When you say continuous derivative, what do you mean? Do you mean that g is differentiable? Also, isn't it a given that the antiderivative of f will be differentiable? – Korgan Rivera Jan 7 '12 at 21:03
I suspected that it wasn't necessary for $f$ to be differentiable, but I wasn't sure since the chain rule requires it to be. I think it's interesting that, after going through the process that arrives at the chain rule, the function $f$ no longer has that constraint when the chain rule itself is manipulated elsewhere, such as the sub. rule. – Korgan Rivera Jan 7 '12 at 21:05
as you can see, we differentiate the composition $Fog$ using chain rule, not $fog$, for proving substitution rule. – NikBels Jan 7 '12 at 21:12