Theorem 1

Let $f$ be a function satisfying $f(c) = 0$ and $\lim\limits_{x \to \infty} f(x)e^{\alpha x} = 0$, for some negative $\alpha$ then we have: $$\int_c^\infty f(x) e^{\alpha x} dx = \frac{1}{-\alpha} \int_c^\infty f'(x) e^{ \alpha x} dx$$ Where $f'(x)$ is the derivative of $f$.

Proof

We'll proceed with integration by parts $$u = f(x)$$ $$du = f'(x) dx$$ $$dv = e^{\alpha x}$$ $$v = \frac{1}{\alpha} e^{\alpha x}$$ $$\int_c^\infty f(x) e^{\alpha x} dx = uv - \int_c^\infty v du$$ $$= f(x) \frac{1}{\alpha} e^{\alpha x} \bigg\rvert_c^\infty - \int_c^\infty f'(x) \frac{1}{\alpha} e^{\alpha x} dx$$ Here we use our conditions $f(c) = 0$ and $\lim\limits_{x \to \infty} f(x)e^{\alpha x} = 0$ and the first term disappears. We pull the constant out front of the integral to end our proof. $$\int_c^\infty f(x) e^{\alpha x} dx = \frac{1}{-\alpha} \int_c^\infty f'(x) e^{ \alpha x} dx$$

There is an analogous theorem and proof for positive $\alpha$.

Theorem 2

Let $f$ be a function satisfying $f(c) = 0$ and $\lim\limits_{x \to -\infty} f(x)e^{\alpha x} = 0$, for some positive $\alpha$ then we have: $$\int_{-\infty}^c f(x) e^{\alpha x} dx = \frac{1}{-\alpha} \int_{-\infty}^c f'(x) e^{ \alpha x} dx$$

Special case

I noticed this property when reading about the Gompertz constant, which has equivalent expressions based off of this property. That particular derivation uses $\alpha = -1$, which is the prettiest: $$\int_c^\infty f(x) e^{-x} dx = \int_c^\infty f'(x) e^{-x} dx$$