I have been working through measure theory, specifically the dominated convergence of Lebesgue integrals and its applications such as differentiating under the integral sign.
There I came across the following example
For $t>0$ it holds $\int_{-\infty}^{\infty} e^{-x^2} \cos(tx) dx = \sqrt{\pi}e^{-t^{2}/4}$
In the solution, I see they first rewrite $\cos(tx) = \lim_{N\to\infty} S_N$, where $S_N=\sum_{n=0}^N \frac{(-1)^n (tx)^{2n}}{(2n)!},$ and then reorganize it to use the dominated convergence theorem as
\begin{align} |e^{-x^2}S_N|&\leq \Big |e^{-x^2}\sum_{n=0}^N \frac{(-1)^n (tx)^{2n}}{(2n)!}\Big|\\ & \leq e^{-x^2} \sum_{n=0}^N \Big|\frac{(-1)^n (tx)^{2n}}{(2n)!}\Big| \\ &\leq e^{-x^2} \sum_{n=0}^N \frac{(tx)^{2n}}{(2n)!}\\ &\leq e^{-x^2} \sum_{n=0}^N \frac{(t_0x)^{2n}}{(2n)!} \\ &=e^{-x^{2}+t_{0} x} \\ &=e^{-(x-\frac{t_{0}}{2})^{2}} e^{\frac{t_{0}^2}{4}}=: g(x) \in L(\mathbb{R}) \end{align}
where $t_0 > 0$ is a fixed number such that $t\in (0,t_0)$.
I don't really see how $e^{-(x-t_0/2)^2} e^{{t_0}^2 /4} := g \in L(\mathbb{R}),$ here $L(X)$ denotes the set of all Lebesgue intergable functions.
The function $g:x\mapsto\exp(-(x-t_0/2)^2)\exp(t_0^2/4)$ lies in $L^1(\Bbb R)$ for each $t_0\in\Bbb R$. We can write $g(x)=A\exp(-(x-B))^2$ where $A=\exp(t_0^2/4)$ and $B=t_0/2$. Since $x\mapsto\exp(-x^2)$ is Lebesgue integrable, so is its translate $x\mapsto\exp(-(x-B)^2)$, and so also the multiple of that, $x\mapsto A\exp(-(x-B)^2)$.