Continuity of the function $x\mapsto \int_0^{+\infty}\frac{e^{-xt}}{1+t^2}dt$ Let the function $\varphi$ be defined by
$$\varphi(x)=\int_0^{+\infty}\frac{e^{-xt}}{1+t^2}dt.$$
$\varphi$ is defined on the interval $[0,+\infty)$ and I want to do a rigorous proof and show that it is continuous on this interval. So, we take $x_0\ge0$, and by the mean value theorem, there is a $c_x$ between $x$ and $x_0$ such that for all $t\ge0$
$$|e^{-xt}-e^{-x_0t}|\le te^{-c_x t}|x-x_0|.$$
Hence we have
$$0\le|\varphi(x)-\varphi(x_0)|\le |x-x_0|\int_0^{+\infty}\frac{te^{-c_xt}}{1+t^2}dt.$$
Now if $x\to x_0$ we have $\varphi(x)\to\varphi(x_0)$.
Is this proof ok? I have a doubt about the limit of the last integral in the inequality because in the case when $c_x\to0$ this integral is no longer convergent. Any suggestions and corrections?
N.B. I know that there is a theorem of continuity for integrals with parameter but I want a simpler proof.
 A: 1. Let $a > 0$ be given. For each $t \geq 0$ and $x, y \in [a, \infty)$, the mean value theorem tells that, for some $\xi$ between $x$ and $y$, we have
$$ |e^{-xt} - e^{-yt}| = |x - y|t e^{-\xi t} \leq |x - y| t e^{-at}. $$
Hence,
$$ \left| \varphi(x) - \varphi(y) \right|
\leq \int_{0}^{\infty} \frac{|e^{-xt} - e^{-yt}|}{1+t^2} \, \mathrm{d}t
\leq |x - y| \int_{0}^{\infty} \frac{t e^{-at}}{1+t^2} \, \mathrm{d}t. $$
This shows that $\varphi$ is Lipschitz continuous (and hence continuous) on $[a, \infty)$ for each $ a > 0 $. Note that this is essentially what OP showed.
2. For each fixed $a > 0$, we have
\begin{align*}
|\varphi(x) - \varphi(0)|
&= \int_{0}^{a} \frac{1 - e^{-xt}}{1 + t^2} \, \mathrm{d}t \\
&\leq (1 - e^{-xa}) \int_{0}^{a} \frac{1}{1 + t^2} \, \mathrm{d}t + \int_{a}^{\infty} \frac{\mathrm{d}t}{1 + t^2}.
\end{align*}
Taking $\limsup$ as $x \to 0^+$, we thus get
$$ \limsup_{x \to 0^+} |\varphi(x) - \varphi(0)|
\leq \int_{a}^{\infty} \frac{\mathrm{d}t}{1 + t^2}. $$
Since the left-hand side does not depend on $a$, letting $a \to \infty$ shows that the indicated limsup is zero, hence proves that $\varphi(x) \to \varphi(0)$ as $x \to 0^+$.
(If you are not fully comfortable with this limsup trick, you may instead plug $a = 1/\sqrt{x}$ to the inequality above and use the squeeze theorem.)

Remark. One can prove that $\varphi(x) = \frac{\pi}{2} + x \log x + \mathcal{O}(x)$ as $x \to 0^+$, hence $\varphi$ is not differentiable at $x = 0$. This explains why the approach in step 1 fails at $x = 0$.
A: Assuming $y>x>0$ we have
$$ \varphi(y)-\varphi(x)=\int_{0}^{+\infty}\frac{e^{-yt}-e^{-xt}}{1+t^2}\,dt .$$
Over $\mathbb{R}^+$ the function $f(z)=e^{-z}$ is decreasing, implying that $e^{-yt}-e^{-xt}$ is negative. By Lagrange's theorem
$$ |e^{-yt}-e^{-xt}| = |y-x|t e^{-zt} \leq |y-x|t e^{-xt} $$
for some $z\in(x,y)$. By Cauchy-Schwarz
$$ \int_{0}^{+\infty}\frac{te^{-xt}}{1+t^2}\,dt\leq\sqrt{\int_{0}^{+\infty}te^{-2xt}\,dt\int_{0}^{+\infty}\frac{t\,dt}{(1+t^2)^2}}=\frac{1}{2x\sqrt{2}}$$
hence it follows that
$$ \left|\varphi(y)-\varphi(x)\right|\leq \frac{|y-x|}{2x\sqrt{2}}. $$
This proves that $\varphi$ is a locally Lipschitz function, hence continuous on $\mathbb{R}^+$. Now we just need to prove the continuity at the origin: for such purpose we may invoke the monotone/dominated convergence theorem, or just exploit simple inequalities. Obviously $\varphi(0)=\frac{\pi}{2}$ and $\varphi$ is non-increasing over $[0,+\infty)$. For any $z>0$
$$\begin{eqnarray*} 0\leq \varphi(0)-\varphi(z) &=& \int_{0}^{+\infty}\frac{1-e^{-zt}}{1+t^2}\,dt\\ &\leq& \int_{0}^{+\infty}\frac{\min(1,zt)}{1+t^2}\,dt=\frac{z}{2}\left(\log(1+z^2)-2\log(z)\right)+\arctan(z)\end{eqnarray*}$$
and luckily $\lim_{z\to 0^+}z\log(z)=0.$
