Differentiating a double integral with respect to its upper limits

In relation to a calculation involving sums of random variables, I needed to differentiate with respect to $\bar{T}$ a double integral of the form

$\int_0^{\bar{T}} \int_0^{\bar{T}-T_1} f(T_1, T_2) \ dT_2 \ dT_1$

The parameter $\bar{T}$ appears in both upper limits, but appears on its own in the outer upper limit, and as $\bar{T} - T_1$ in the inner upper limit. I went through the motions of using Liebniz’s Rule and was interested to see how it worked in this case. I have not seen anything like this anywhere else online, so I want to record it here.

Liebniz’s Rule in the case of a single integral works thus:

$\frac{d}{dx} \int_0^x f(t) dt = \frac{d}{dx} \big[F(x) - F(0)\big] = f(x)$

Here, $F$ is the antiderivative of $f$ . The trick in the double integral case is to treat the inner integral like the function in the single integral case above. So, we differentiate the outer integral treating the inner integral as the function in the single integral case, then we differentiate the inner integral in the normal way. Thus, we write

$\frac{d}{d\bar{T}} \int_0^{\bar{T}} \int_0^{\bar{T}-T_1} f(T_1, T_2) \ dT_2 \ dT_1$

$= \frac{d}{d\bar{T}}\int_0^{\bar{T}} \bigg(\int_0^{\bar{T}-T_1} f(T_1, T_2) \ dT_2 \bigg) \ dT_1 + \int_0^{\bar{T}} \frac{d}{d\bar{T}} \bigg( \int_0^{\bar{T}-T_1} f(T_1, T_2) \ dT_2\bigg) \ dT_1$

$= \int_0^{\bar{T} - \bar{T}} f(\bar{T}, T_2) \ dT_2 + \int_0^{\bar{T}} f(T_1, \bar{T} - T_1) \ dT_1$

$= \int_0^{\bar{T}} f(T_1, \bar{T} - T_1) \ dT_1$

The first integral in the penultimate step vanishes because the upper limit is converted from $\bar{T} - T_1$ into $\bar{T} - \bar{T}$ in the application of Liebniz’s Rule.

Differentiating a double integral with respect to its upper limits

Published by Dr Christian P. H. Salas

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