## 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 is a function on the rectangle , and that is of bounded variation on and that the integral

exists for every . Further, assume that the partial derivative exists and is continuous throughout . Then the derivative of exists and is given by

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

So, if and we can calculate the difference quotient:

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

as we take to approach , the number gets squeezed towards as well. Since we assumed to be continuous, the limit in the integrand will equal , as asserted.

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 |

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

Comment by John Armstrong | January 13, 2010 |

Yes, but for each the value between and may be different. Right?

Comment by Nick | January 13, 2010 |

Oh, yes. depends on , but when we take the limit it gets squeezed against at every point.

Comment by John Armstrong | January 13, 2010 |

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 (, and ), 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 |

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

Comment by John Armstrong | January 14, 2010 |

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

Comment by peeterjoot | January 14, 2010 |

Oh, may well fail to exist even if

iscontinuous.Comment by John Armstrong | January 14, 2010 |

Even when exists? You require more than continuity for that (like the example function in equation (1))

http://books.google.ca/books?id=jlQnThDV41UC&lpg=PP1&dq=functional%20analysis%20nagy&pg=PA4#v=onepage&q=&f=false

That function was continuous but nowhere differentiable here (graphing that function was about as far as I could ever get in that book;)

Comment by peeterjoot | January 14, 2010 |

Just go back to the definition of continuity and go from there.

As for Riesz/Nagy.. I taught myself French math reading that one in both languages. Good times. Not the most readable treatment, though.

Comment by John Armstrong | January 14, 2010 |

[...] 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 |

John – I have a rather technical issue with your proof. You assume that 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 isn’t even well defined by this argument (although, I’ve no doubt that 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 |

Actually, maybe I spoke too soon. You already know that is continuous, because it is equal to the finite difference . We don’t really care what is, and could be unmeasurable for all it matters.

Comment by George Lowther | January 15, 2010 |