I have to study punctual and uniform convergence of $$g(x):=\sum_{n=1}^{\infty}\int_n^{n+x}\dfrac{dt}{e^{t^3}}.$$ If I define $g_n(x)=\int_n^{n+x}\frac{dt}{e^{t^3}}$, in order to study the punctual convergence of $g(x)$, I have to compute $\lim_{n\to \infty}g_n$ with $x\in \mathbb R$ fixed. Since for $n\to \infty$, $n+x\sim n$ $\forall x\in \mathbb R$ I wrote $$\int_n^{n+x}\dfrac{dt}{e^{t^3}}=\int_0^{n+x}\dfrac{dt}{e^{t^3}}-\int_0^{n}\dfrac{dt}{e^{t^3}}$$ which (I think) tends to zero and I'd say that the limit function of $g_n$ is the zero function, but I'm not sure about this. If I want to deduce the convergence of the series I should find a function asymptotic to $g_n$. The sum has form $\int_1^{1+x}\frac{dt}{e^{t^3}}+\int_2^{2+x}\frac{dt}{e^{t^3}}+\dots$, so how could I get informations about the convergence of the series? I don't understand how to do with the $n+x$ term because for $n<\infty$ it can assume any real value.
I thank you in advance for your help.
It converges uniformly on any set $[a; +\infty)$, but doesn't converges uniformly on $\mathbb R$.
It's enough to check case $a < 0$. If $n > -a + 1$, $|g_n(x)| = \int\limits_{n + a}^n \frac{dt}{\exp(t^3)} < \int\limits_{n + a}^n e^{-t}\,dt < e^{-n - a}$. Taking $M_n = e^{-n - a}$ note that $\sum\limits_{n=0}^\infty M_n$ converges, thus $\sum g_n(x)$ converges uniformly by Weierstrass M-test.
It doesn't converge uniformly on $\mathbb R$, because $g_n(-n - 1)$ doesn't even converges to $0$.