Calculate $\lim_{j \rightarrow \infty} \int_0^j (1+\frac{x}{j})^j e^{-\pi x}dx$

Question

My approach: $$\lim_{j \rightarrow \infty} \int_0^j \left(1+\frac{x}{j}\right)^j e^{-\pi x}dx = \lim_{j \rightarrow \infty} \int_0^{\infty} \left(1+\frac{x}{j}\right)^j e^{-\pi x}dx.$$ I'm not sure if I can do the above, but I need it to apply Lebesgue as follows:

I see that $(1+\frac{x}{j})^j \rightarrow e^x$ and $(1+\frac{x}{j})^j$ is monotonic increasing, so $e^x$ is a dominating function that is integrable. Hence I can apply Lebesgue:

\begin{align*} \lim_{j \rightarrow \infty} \int_0^{\infty} \left(1+\frac{x}{j}\right)^j e^{-\pi x} dx & = \int_0^{\infty} \lim_{j \rightarrow \infty} \left(1+\frac{x}{j}\right)^j e^{-\pi x}dx \\ &= \int_0^{\infty} e^x e^{-\pi x} dx= \int_0^{\infty} e^{(1-\pi) x} dx= - \frac{1}{1-\pi}. \end{align*} My colleagues all have a different result though and Wolfram Alpha exceeds computation time. I cannot spot a mistake.

Answer 1

Your answer is correct, but first statement $$ \lim_{j \rightarrow \infty} \int_0^j \left(1+\frac{x}{j}\right)^j e^{-\pi x} dx = \lim_{j \rightarrow \infty} \int_0^{\infty} \left(1+\frac{x}{j}\right)^j e^{-\pi x} dx $$ needs some clarification: $$ \int_0^j \left(1+\frac{x}{j}\right)^j e^{-\pi x} dx = \int_0^{+\infty}1_{[0, j]}(x)\left(1+\frac{x}{j}\right)^j e^{-\pi x} dx $$ where $1_{[0, j]}(x)$ is the indicator function of the set $[0, j]$. Now you can apply dominated convergence theorem to $1_{[0, j]}(x)\left(1+\frac{x}{j}\right)^j e^{-\pi x}$.