Is this integral continuous? (with respect to $z$)

Question

Consider the integral $$\int_0^\infty f(t)e^{tz}\,dt,$$ where $f$ is an integrable function.

Is this integral continuous with respect to $z$ (complex variable) on the domain $\{z=x+yi:x<0,y\in\mathbb{R}\}$?

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Is this proof acceptable?

Since $|f(t)e^{tz}|\leq f(t)$, and $f$ is integrable, so by the Dominated Convergence Theorem (continuous version),

$$\lim_{z\to z_0}\int_0^\infty f(t)e^{tz}\,dt=\int_0^\infty f(t)e^{tz_0}\,dt$$

Thus, the integral is continuous (wrt $z$).

Thanks!

Answer 1

Yes, this works, except that you don't have $$ |f(t)e^{tz}|\leq f(t) $$ but instead $$ |f(t)e^{tz}|\leq |f(t)| \tag 1 $$ which also works since you assumed $f(t)$ to be integrable. By $(1)$ it follows by using the dominated convergence theorem, that $f(t)e^{tz}$ is integrable as well and that you can exchange the limit with the integral. And therefore $$ F(z)=\int_0^\infty f(t)e^{tz}\,dt $$ is continuous in $z$.