Uniform Convergence of Improper Integral

Question

Define $$F(t) = \int_{-\infty}^{\infty}\frac{x+t}{e^{x^2}+e^t}dx$$ It seems to me that this integral converges uniformly for all $t\in[A,+\infty)$, $A\in\mathbb{R}$, since $e^t$ grows much faster than $t$. However, I cannot find an appropriate inequality tho show this... I only have: $$\bigg|\frac{x+t}{e^{x^2}+e^t}\bigg|\leq\frac{|x|+|t|}{e^{x^2}+e^t}\leq\frac{|x|}{e^{x^2}}+\frac{|t|}{e^{x^2}+e^t}$$ Evidently the first term converges, but I've tried many inequalities with the second one and nothing seems to work... I checked it on wolfram and it's apparently smaller than $1/x^2$, if I can prove that then I'd be done, but I don't see how to prove it.

Answer 1

Convergence is uniform by the Weierstrass test -- $\int_\mathbb{R} f(x,t) \,dx$ is uniformly convergent if there exists a uniform bounding function $g$ with $|f(x,t)| \leqslant g(x)$ such that $\int_\mathbb{R} g(x) \, dx < \infty$.

The first term of your bound $|x|/ e^{x^2}$is integrable over $\mathbb{R}$. For the second term of the bound, we have for $t \in [0,\infty)$,

$$\frac{|t|}{e^{x^2} + e^t} < \frac{t}{e^{x^2}+t^2} \leqslant \max_{t \in [0,\infty)}\frac{t}{e^{x^2}+t^2} = \frac{\sqrt{e^{x^2}}}{2e^{x^2}} =\frac{1}{2e^{x^2/2}}, $$

and, in case $A < 0$, for $t \in [A,0]$ we have

$$\frac{|t|}{e^{x^2} + e^t} \leqslant \frac{|A|}{e^{x^2}}.$$

In either case, the RHS is integrable and we can conclude that convergence of the integral is uniform.