Can we do anything with $\frac{d}{db}\int_0^1e^{bx}f(x)dx$?

Question

I have this monster as the part of a longer calculation. My goal would be to somehow make it... nicer.

Intuitively, I would try to somehow utilize the derivate and the integral against each other, but I have no idea, exactly how.

I suspect, this might be a relative common problem, i.e. if we want to derivate a parametrized definite integral against one of its parameters. Maybe it has even some trick... or methodology to handle it.

Of course it is not a problem if the integral remains. The primary goal would be in this stage, to eliminate or significantly simplify the derivative operator.

Answer 1

This is probably in your calculus textbook ... under reasonable conditions,

$$ \frac{d}{db}\int_0^1e^{bx}f(x)dx = \int_0^1 \frac{\partial}{\partial b}e^{bx}f(x)dx = \int_0^1 x e^{bx}f(x)dx $$

Answer 2

If $f$ is continuous you can just use the Leibniz integral rule to assert: $\frac{d}{db}\int_0^1 e^{bx} f(x)dx = \int_0^1 x e^{bx} f(x)dx$

The Leibniz integral rule says that if $g$ and its partial derivatives are continuous, then $$ \frac{d}{d y} \int_a^b g(x,y)dx = \int_a^b \frac{\partial}{\partial y} g(x,y)dx $$

Answer 3

I looked for a way to simplify such things, and found out that trading differentiation for integration was an efficient way to simplify the task. In your case, \begin{align} \int_{0}^{1}e^{bx}f(x)dx &= \int_{0}^{1}\left(\int_{0}^{b}xe^{cx}dc+1\right) f(x)dx\\ & = \int_{0}^{b}\int_{0}^{1}e^{cx}xf(x)dx dc+\int_{0}^{1}f(x)dx. \end{align} The function $c\mapsto \int_{0}^{1}e^{cx}xf(x)dx$ is a continuous function of $c$. So the Fundamental Theorem of Calculus now implies that the left side is differentiable in $b$ because the right side is, and $$ \frac{d}{db}\int_{0}^{1}e^{bx}f(x)dx = \int_{0}^{1}e^{bx}xf(x)dx. $$ It turns out that trading differentiation for integration in this way is a handy way to justify such operations. Continuity in $c$ of the function mentioned above also follows from integral theorems.