# Differentiation under the integral sign – Leibniz integral rule

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Let us say we have a function defined by a definite integral, and that the function depends on some parameter, p. From the definition of an Riemann Integral we have

$I(p) = \int\limits_a^b {f(x,p)dx} = \mathop {\lim }\limits_{\Delta x \to 0} \sum\limits_{n = 0}^{N - 1} {f(x,p)\Delta x}$

Let us now take the derivative of this with respect to p, what would we get? (We shall here approach the problem in a somewhat semi-formal way. For a more formal proof you may google for Leibniz integral rule.) We get

$I'(p) = \frac{d}{{dp}}I(p) = \mathop {\lim }\limits_{h \to 0} \frac{{I(p + h) - I(p)}}{h}$

or

$I'(p) = \mathop {\lim }\limits_{h \to 0} \frac{{\int\limits_a^b {f(x,p + h)dx} - \int\limits_a^b {f(x,p)dx} }}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\int\limits_a^b {(f(x,p + h) - f(x,p))dx} }}{h}$

So

$I'(p) = \mathop {\lim }\limits_{h \to 0} \frac{{\mathop {\lim }\limits_{\Delta x \to 0} \sum\limits_{n = 0}^{N - 1} {(f(x,p + h) - f(x,h))\Delta x} }}{h} = \mathop {\lim }\limits_{h \to 0} \mathop {\lim }\limits_{\Delta x \to 0} \sum\limits_{n = 0}^{N - 1} {\frac{{(f(x,p + h) - f(x,h))\Delta x}}{h}}$

The last step was done because since h is a constant, as seen from the inner limit, we may put it inside the sum (this has to be proven – but for now I’ll let it pass).

Now, since h and Δx are independent, we may switch the limits, to get

$I'(p) = \mathop {\lim }\limits_{\Delta x \to 0} \mathop {\lim }\limits_{h \to 0} \sum\limits_{n = 0}^{N - 1} {\frac{{(f(x,p + h) - f(x,h))\Delta x}}{h}}$

This is yet another step that may need some more rigorous work – but it will do for now. In this we recognize the definition of derivative, and we may thus write

$I'(p) = \mathop {\lim }\limits_{\Delta x \to 0} \sum\limits_{n = 0}^{N - 1} {\frac{\partial }{{\partial p}}f(x,p)\Delta x}$

or

$\frac{d}{{dp}}\int\limits_a^b {f(x,p)dx} = \int\limits_a^b {\frac{\partial }{{\partial p}}f(x,p)dx}$

Ok, so we may switch the order – how would that help us? I mean, we don’t get the value of the integral – right?

In the next few sections, we will look at some examples of how this could be used.

Up a level : Integrals
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Next page : The integral of (e^(-px)-e^(-qx))/x dx from 0 to infinityLast modified: Dec 28, 2023 @ 14:46