In this post the author uses differentiation under the integral sign to evaluate the integral

$$ \int_0^1\frac{\log(1+x)}{1+x^2}\,\mathrm dx $$

by setting up the function

$$ F(a) = \int_0^1\frac{\log(1+ax)}{1+x^2}\,\mathrm dx, $$

and computing $F'(a) = \int_0^1\frac{\partial}{\partial a}\big[\frac{\log(1+ax)}{1+x^2}\big]\,\mathrm dx$. My question is how do we formally justify the use of differentiation under the integral sign here?

Some thoughts. Let's put $f(x,a) = \frac{\log(1+ax)}{1+x^2}$ for $x\in[0,1]$ and $a$ in some domain $A\subseteq\mathbf R$. I know that sufficient conditions for this procedure are that $f(x,a)$ and $\frac{\partial f}{\partial a}(x,a)$ are continuous on $[0,1]\times A$, that $F(a)$ exists for each $x\in[0,1]$, and that for each $a\in A$, $\big|\frac{\partial f}{\partial a}(x,a)\big|\le g(x)$ where $\int_0^1g(x)\,\mathrm dx<\infty$. I'm not sure how to use these ideas to put together a proof, and maybe we don't need each of these ideas to do it.

So how do we justify this?

  • $\begingroup$ I'm not sure what you're asking: the first half of your question seems to be "how do we justify this," but the second half seems to be "we're justifying this with these things, but do we need them all?" $\endgroup$ – Adam Hughes Dec 30 '16 at 22:33
  • $\begingroup$ @AdamHughes It's one question: How do we formally justify the procedure? I just posted some of my thoughts, nothing like a complete proof. $\endgroup$ – Alex Ortiz Dec 30 '16 at 22:38
  • $\begingroup$ Are you unclear on why differentiation under the integral sign works at all, or why it works in this case? Why might it not work? $f(a,x)$ is continuous and defined (at least while $a\ge 0$ and $x\in (0,1]$) same for $\frac {\partial f}{\partial a}$. Where is your doubt? $\endgroup$ – Doug M Dec 30 '16 at 23:15

Your "thoughts" are on the right track. For this class of problems, we rely on the compact set of the domain of the function and its derivative, which renders the function and its derivative uniformly continuous.

If $a$ is restricted to the closed interval $[a_1,a_2]$, with $a_1>-1$, then $f(x,a)=\frac{\log(1+ax)}{1+x^2}$ and $\frac{\partial f(x,a)}{\partial a}$ are uniformly continuous on $[0,1]\times [a_1,a_2]$.

Let $\epsilon>0$ be given, and choose $\delta>0$ such that $|x-x'|<\delta$ and $|a-a'|<\delta$ imply $|\frac{\partial f(x,a)}{\partial a}-\frac{\partial f(x',a')}{\partial a'}|<\epsilon$. Next, choose $0<|h|\le \delta$ and let $a\in [a_1,a_2]$.

Then, from the mean-value theorem, there exists a number $0<\theta(x)<1$ such that

$$\begin{align} \left|\int_0^1 \left(\frac{f(x,a+h)-f(x,a)}{h}-\frac{\partial f(x,a)}{\partial a}\right)\,dx\right|&=\left|\int_0^1 \left(\frac{\partial f(x,a+\theta h)}{\partial a}-\frac{\partial f(x,a)}{\partial a}\right)\,dx\right|\\\\ &\le \int_0^1 \left|\frac{\partial f(x,a+\theta h)}{\partial a}-\frac{\partial f(x,a)}{\partial a}\right|\,dx\\\\ &<\epsilon \end{align}$$

And we are done!

Note that although we don't understand the nature of $\theta(x)$, $\frac{\partial f(x,a+\theta h)}{\partial a}$ is an integrable function since it is equal to the difference quotient $\frac{f(x,a+h)-f(x,a)}{h}$.


This is an application of a specific case of the Leibniz integral rule in the case where the bounds don't depend on the variable of differentiation. It goes something like this: Given that f(x, t) is a function such that the partial derivative of f with respect to t exists, and is continuous, then

\begin{align} {\frac {\mathrm {d} }{\mathrm {d} t}}\left(\int _{a(t)}^{b(t)}f(x,t)\,\mathrm {d} x\right)=\int _{a(t)}^{b(t)}{\frac {\partial f}{\partial t}}\,\mathrm {d} x\,+\,f{\big (}b(t),t{\big )}\cdot b'(t)\,-\,f{\big (}a(t),t{\big )}\cdot a'(t) \end{align}

There are plenty of proofs of this. My favorite is in the book called $Inside\ Interesting\ Integrals$, in which the author just uses the familiar limit definition of the partial derivative to prove it. It follows almost directly from those definitions.

  • 1
    $\begingroup$ I don't really find this answer sufficient considering what I am looking for is an actual proof that the procedure works in this case. Maybe the answer could be reworded and written like an actual proof, or someone could point out to me how this actually does suffice as an acceptable answer. $\endgroup$ – Alex Ortiz Dec 30 '16 at 22:47
  • $\begingroup$ @AOrtiz What's not to like of this? $\endgroup$ – Simply Beautiful Art Dec 30 '16 at 23:18
  • $\begingroup$ @SimpleArt The OP's comment states ... "what I am looking for is an actual proof that the procedure works in this case." $\endgroup$ – Mark Viola Dec 31 '16 at 3:14

Your thoughts are essentially from a theorem (2.27b) in Folland's Real Analysis:

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A simple calculation shows that $$ \biggr|\frac{\partial}{\partial a}\left[\frac{\log(1+ax)}{1+x^2}\right]\biggr|= \biggr|\frac{x}{1+x^2}\frac{1}{1+ax}\biggr|\leq 1 $$ for $(x,a)\in[0,1]\times[0.5,1.5]$. Thus one can let $g(x)=1$ on $[0,1]$ as the dominated function.

  • $\begingroup$ The OP's comment states ... "what I am looking for is an actual proof that the procedure works in this case." $\endgroup$ – Mark Viola Dec 31 '16 at 3:15

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