Evaluating $\int_0^{\infty}\frac{e^{-x}}{1+x^2}dx$

As pointed out in the comments and in Xiaolang's answer, this integral is very unlikely to have a closed-form solution in terms of elementary functions, but it is possible to express it in terms of exponential integrals or, equivalently, sine and cosine integrals. These hardly qualify as "elementary methods", but we can at least eliminate the sine and cosine integrals using their power series expansions.

We start with the representation in terms of the cosine and sine integrals $\operatorname{Ci}$ and $\operatorname{Si}$: \begin{align} \int_0^\infty \frac{e^{-x}}{1+x^2} &= \operatorname{Ci}(1)\sin(1) - \operatorname{Si}(1)\cos(1)+\frac{\pi \cos(1)}{2},\\\ \\ \operatorname{Ci}(x) &= \gamma + \ln(x) +\int_0^x \frac{\cos(t)-1}{t}dt, \\\ \\ \operatorname{Si}(x) &= \int_0^x \frac{\sin(t)}{t}dt, \end{align} where $\gamma=0.577\ldots$ is the Euler-Mascheroni constant. Power series expansions for $\operatorname{Ci}(x)$ and $\operatorname{Si}(x)$ follow directly from their definitions in terms of sines and cosines: \begin{align} \operatorname{Ci}(x) &= \gamma + \ln(x) + \sum_{j=0}^\infty \frac{(-1)^{j+1}x^{2j+2}}{(2j+2)!(2j+2)},\\\ \\ \operatorname{Si}(x) &= \sum_{j=0}^\infty \frac{(-1)^jx^{2j+1}}{(2j+1)!(2j+1)}. \end{align} Substituting $x=1$ into these expressions, applying the power series expansions for $\sin(1)$ and $\cos(1)$ where they multiply $\operatorname{Ci}(1)$ and $\operatorname{Si}(1)$, and doing a somewhat non-negligible amount of arithmetic, we find: $$\int_0^\infty \frac{e^{-x}}{1+x^2} = \gamma \sin(1) + \frac{\pi\cos(1)}{2} + \sum_{k=0}^\infty \sum_{j=1}^\infty \frac{f(j-k)}{j(j!)(k!)}. \tag{$\star$}$$ Here the function $f(j-k)$ is the sequence $0,-1,0,1,\ldots$: \begin{align} f(j-k) &= 0\ \ \quad \textrm{if}\quad j-k\equiv 0\ (\operatorname{mod} 4)\\ &= -1\ \ \ \,\textrm{if}\quad j-k\equiv 1\ (\operatorname{mod} 4)\\ &= 0\ \ \quad \textrm{if}\quad j-k\equiv 2\ (\operatorname{mod} 4)\\ &= 1\ \ \quad \textrm{if}\quad j-k\equiv 3\ (\operatorname{mod} 4).\\ \end{align} Clearly Eq. $(\star)$ converges rapidly, considering the factorial functions in the denominator of the summand. Indeed, including just 25 terms (5 values for each index), we obtain 0.621..., consistent with the result 0.621449... quoted previously.

there may not be a elementray functions but it isn't a easy thing to prove you can consult The Liouville's therom in complex analysis

but you can get a approximation (WolframAlpha can help you)

$$\int_0^{\pi/2}e^{\tan(\theta)}d\theta= \operatorname{Ci}(1)\sin(1)- \operatorname{Si}(1)\cos(1)+\pi \cos(1)/2 \approx 0.62144962423581335763926$$