Calculate the improper integral and the Taylor series of $f(x) = \int_0^\infty {e^{-t}\over 1+x \cdot t} \,\mathrm dt$

Question

For the given function: $$f(x) = \int_0^\infty {e^{-t}\over 1+x \cdot t} \,\mathrm dt$$ with $x>0$, calculate the Taylor series of $f(x)$ at $x=0$.

I tried different stuff, but I did not get very far. I would really appreciate any help for solution.

Answer 1

By differentiating under the integral sign, $$ f^{(n)}(0) = (-1)^n n!\int_{0}^{+\infty}t^n e^{-t}\,dt = (-1)^n n!^2$$ hence the Taylor series of $f(x)$ in a neighbourhood of the origin is given by: $$ f(x) = \sum_{n\geq 0}(-1)^n n!\, z^n. $$

Answer 2

We have $$f(x) = \int_0^\infty e^{-t}\sum_{n = 0}^\infty (-1)^n x^n t^n\, dt = \sum_{n = 0}^\infty (-1)^n x^n \int_0^\infty e^{-t} t^n\, dt = \sum_{n = 0}^\infty (-1)^n n! x^n.$$