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This question is concerned with the integral with parameter, so let's assume that every function below is smooth.

To find the formula for the derivative of an integral with parameter, say $$g(x) = \int_{a(x)}^{b(x)}f(x,t)dt$$ for some function $f: \mathbb{R}^2 \to \mathbb{R}$, one would define $$ F: \mathbb{R}^3 \to \mathbb{R}, F(a,b,x) = \int_a^b f(x,t)dt $$ and then use the multivariable chain-rule to determine $g'$.

Thus, we would have that $$g'(x) = \frac{\partial F}{\partial a} \cdot \frac{\partial a}{\partial x} + \frac{\partial F}{\partial b} \cdot \frac{\partial b}{\partial x} + \frac{\partial F}{\partial x}.$$

It is easy so see that $$\frac{\partial F}{\partial a} \cdot \frac{\partial a}{\partial x} = -f(x,a(x)) \cdot a'(x)$$ and that $$\frac{\partial F}{\partial b} \cdot \frac{\partial b}{\partial x} = f(x,b(x)) \cdot b'(x). $$

What I don't understand is why we have that $$\frac{\partial F}{\partial x} = \int_a^b \frac{\partial f}{\partial x}(x,t)dt. $$

Shouldn't it be $$\frac{\partial F}{\partial x} = \frac{\partial}{\partial x}\int_a^b f(x,t)dt ? $$

Could you please tell me why we are allowed to interchange the partial derivative and the integral?

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  • By the way, this is the Leibniz rule (see https://en.wikipedia.org/wiki/Leibniz_integral_rule). – Minus One-Twelfth Mar 18 '19 at 10:45
  • I don't know your background (e.g. do you know any measure theory) so I don't know which related question question to link you to, but this only works under certain conditions and is called "differentiation under the integral sign". See, for instance, https://math.stackexchange.com/q/246380/26369 – Mark S. Mar 18 '19 at 10:46
  • This is because $a$ and $b$ are assumed constant during differentiation with respect to $x$... – Mostafa Ayaz Mar 18 '19 at 11:47

1 Answers1

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$$\frac{\partial F}{\partial x} = \int_a^b \frac{\partial f}{\partial x}(x,t)dt$$

and

$$\frac{\partial F}{\partial x} = \frac{\partial}{\partial x}\int_a^b f(x,t)dt $$

are essentially the same thing in this context, although it's more often referred to as

$$\frac{\partial F}{\partial x} = \frac{d}{d x}\int_a^b f(x,t)dt= \int_a^b\frac{\partial}{\partial x} f(x,t)dt ?$$

As @MinusOne-Twelfth pointed out this is the Leibniz rule.

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