Define $$I(t) := \int_0^\infty e^{-x^2} \cosh(tx) dx , \quad J(t) := \int_0^\infty xe^{-x^2}\sinh(tx) dx.$$
question
(1)Both $I(t)$ and $J(t)$ uniformly converge on $[0, R)\quad (R>0).$
(2)Calculate $I(t)$.
attempt (1)I know the definition of uniform convergence of improper integral. I tried finding $g$ that satisfies $(i)e^{-x^2} \cosh(tx) \leq g(x)$, and $(ii) g $ is improper integrable. But I could not, because $\cosh(tx)$ grows forever.
(2)I could calculate $I(0)$, because $I(0) = \int_0^\infty e^{-x^2} dx$. And I think integration by parts repetition may help.
edit
I understand (2).
It holds
$$I(t) = \frac{1}{2}I^\prime(t),$$
so the rest is easy.