How to Compute: $\int_\mathbb{R} \frac{\lambda e^{-t}}{(\lambda - e^{-t})^2}f(x)\chi_{(-\infty,t)}(x) \, dt$?

Question

Let $\lambda \in \mathbb{C}\backslash\mathbb{R}_+$ and $f \in L^2$. Then

$$ (\lambda - e^{-x})^{-1}f(x) = \int_\mathbb{R} \frac{\lambda e^{-t}}{(\lambda - e^{-t})^2}f(x)\chi_{(-\infty,t)}(x) \, dt $$

for any $x \in \mathbb{R}$.

How could one derive the above equality?

Answer 1

Note that $$\color{red}{\chi_{(-\infty,t)}(x) =\chi_{(x,\infty)}(t)}$$

\begin{align}\int_\mathbb{R} \frac{\lambda e^{-t}}{(\lambda - e^{-t})^2}f(x)\color{red}{\chi_{(-\infty,t)}(x)} \, dt &= \int_\mathbb{R} \frac{\lambda e^{-t}}{(\lambda - e^{-t})^2}f(x)\color{red}{\chi_{(x,\infty)}(t) }\, dt\\ &=f(x)\int_x^{\infty} \frac{\lambda e^{-t}}{(\lambda - e^{-t})^2} \, dt \\&= f(x)\int_x^{\infty} \frac{\lambda e^{t}}{(\lambda e^{t}-1)^2} \, dt \\&=f(x)\int_x^{\infty}\left( -\frac{1}{(\lambda e^{t}-1)}\right)' \, dt \\&=\frac{f(x)}{(\lambda e^{x}-1)}\\&=e^{-x}(\lambda - e^{-x})^{-1}f(x) \end{align}