# The Unapologetic Mathematician

## Differentiation Under the Integral Sign

Another question about “partial integrals” is when we can interchange integral and differential operations. Just like we found for continuity of partial integrals, this involves interchanging two limit processes.

Specifically, let’s again assume that $f:R\rightarrow\mathbb{R}$ is a function on the rectangle $R=[a,b]\times[c,d]$, and that $\alpha$ is of bounded variation on $[a,b]$ and that the integral

$\displaystyle F(y)=\int\limits_a^bf(x,y)\,d\alpha(x)$

exists for every $y\in[c,d]$. Further, assume that the partial derivative $\left[D_2f\right](x,y)$ exists and is continuous throughout $R$. Then the derivative of $F$ exists and is given by

$\displaystyle\frac{d}{dy}F(y)=\frac{d}{dy}\int\limits_a^bf(x,y)\,d\alpha(x)=\int\limits_a^b\frac{\partial}{\partial y}f(x,y)\,d\alpha(x)$

Similar results hold where $x$ and $y$ are vector values, and where derivatives in terms of $y$ are replaced outside the integral by partial derivatives in terms of its components.

So, if $y_0\in(c,d)$ and $y\neq y_0$ we can calculate the difference quotient:

\displaystyle\begin{aligned}\frac{F(y)-F(y_0)}{y-y_0}&=\int\limits_a^b\frac{f(x,y)-f(x,y_0)}{y-y_0}\,d\alpha(x)\\&=\int\limits_a^b\left[D_2f\right](x,\xi)\,d\alpha(x)\end{aligned}

where $\xi$ is some number between $y_0$ and $y$ that exists by the differential mean value theorem. Now to find the derivative, we take the limit of the difference quotient

\displaystyle\begin{aligned}\frac{d}{dy}F(y_0)&=\lim\limits_{y\rightarrow y_0}\frac{F(y)-F(y_0)}{y-y_0}\\&=\lim\limits_{y\rightarrow y_0}\int\limits_a^b\left[D_2f\right](x,\xi)\,d\alpha(x)\\&=\int\limits_a^b\lim\limits_{y\rightarrow y_0}\left[D_2f\right](x,\xi)\,d\alpha(x)\end{aligned}

as we take $y$ to approach $y_0$, the number $\xi$ gets squeezed towards $y_0$ as well. Since we assumed $\left[D_2f\right](x,y)$ to be continuous, the limit in the integrand will equal $\left[D_2f\right](x,y_0)$, as asserted.

January 13, 2010 - Posted by | Analysis, Calculus

1. Doesn’t the differential MVT only imply that for each x there is a \xi(x) between y and y_0 such that…? It looks here that you’re applying the MVT to F(y) if we already know what F'(y) is, but I may be missing something.

Comment by Nick | January 13, 2010 | Reply

2. Nick, I’m using the mean value theorem here for $f$, not for $F$. Remember that the partial derivative $D_2f$ is just like a regular derivative if we hold $x$ fixed, and so the differential mean value theorem applies.

Comment by John Armstrong | January 13, 2010 | Reply

3. Yes, but for each $x$ the value $\xi$ between $y$ and $y_0$ may be different. Right?

Comment by Nick | January 13, 2010 | Reply

4. Oh, yes. $\xi$ depends on $x$, but when we take the limit it gets squeezed against $y_0$ at every point.

Comment by John Armstrong | January 13, 2010 | Reply

5. As somebody trained in an engineering school, the temptation is to consider this obvious and skip right to the end without justification. It was enjoyable to see how the mean value theorem could be used to avoid such a handwaving argument.

Continuity of f(x,y) when y is varied appears to be required. I am guessing this is a requirement of the mean value theorem? Is there also a continuity requirement for x variation of f?

To extend this argument to the case where the boundaries are functions of y ($b = b(y)$, and $a = a(y)$), I suppose you’d decompose the integral into sums. How to procede from there is not entirely clear to me.

Comment by peeterjoot | January 14, 2010 | Reply

6. Continuity with respect to $y$ is required in order to define $D_2f$ in the first place. As for continuity in $x$, I’ll put it to you as a question: can you derive it from the assumption that $D_2f$ is continuous?

Comment by John Armstrong | January 14, 2010 | Reply

• Thanks. I think I see what you are saying. Continuity in a neighbourhood is required for the $D_2 f$, which will mean that $D_1 f$ also exists.

Comment by peeterjoot | January 14, 2010 | Reply

• Oh, $D_1f$ may well fail to exist even if $f$ is continuous.

Comment by John Armstrong | January 14, 2010 | Reply

7. […] Partial Integrals A neat little variation on differentiating an integral from last time combines it with the fundamental theorem of calculus. It’s especially […]

Pingback by Differentiating Partial Integrals « The Unapologetic Mathematician | January 14, 2010 | Reply

8. John – I have a rather technical issue with your proof. You assume that $\xi$ is a function of x. It is not known to be a continuous function of x, and the mean value theorem only tells you that it exists, without giving you any properties of the function. Therefore, the integral $\int D_2f(x,\xi)dx$ isn’t even well defined by this argument (although, I’ve no doubt that $\xi$ will indeed be measurable).

I think you only need to show that $\latex (f(x,y)-f(x,y_0))/(y-y_0)\to D_2 (f(x,y_0)$ uniformly, which the MVT does prove, and this limit can be taken inside the integral.

Comment by George Lowther | January 15, 2010 | Reply

• Actually, maybe I spoke too soon. You already know that $D_2f(x,\xi)$ is continuous, because it is equal to the finite difference $(f(x,y)-f(x,y_0))/(y-y_0)$. We don’t really care what $\xi$ is, and could be unmeasurable for all it matters.

Comment by George Lowther | January 15, 2010 | Reply

9. Does it still hold when $\latex g_y(x)=f(x,y)$ is singular at integration limits but still integrable?

Comment by a naive user | July 29, 2015 | Reply

• I don’t know offhand. I’d suggest looking up the book Counterexamples in Analysis, as they’d likely have a good one if there is one.

Comment by John Armstrong | August 2, 2015 | Reply