How may we show that $\int_{0}^{\pi/2}e^{-{\pi\over 2}\tan t}\mathrm dt=C_i\left({\pi\over 2}\right)?$ Proposed:

$$\int_{0}^{\pi/2}e^{-{\pi\over 2}\tan t}\mathrm dt=C_i\left({\pi\over 2}\right)\tag1$$

Where $C_i$ is Cosine Integral
$$C_i(x)=-\int_{x}^{\infty}{\cos t\over t}\mathrm dt\tag2$$
My try:
Recall (I doubt it would be any useful) $$e^{\tan t}=1+t+{t^2\over 2}+{t^3\over 2}+{3t^3\over 8}+\cdots\tag3$$
$u={\pi\over 2}\tan t\implies {\pi\over 2}\sec ^2 t$ then $(1)$ becomes
$$2\pi\int_{0}^{\infty}e^{-u}\cdot{\mathrm du\over 4u^2+\pi^2}\tag4$$
Recall $$\int{\mathrm du\over u^2+a^2}={1\over a}\tan^{-1}{u\over a}\tag5$$
probably $(4)$ we may apply integration by parts?
How does one prove $(1)$?
 A: Define
$$
F(\alpha)=\int_0^{\pi/2}e^{-\alpha\tan x}\,dx,\quad\alpha>0.
$$
Differentiating $F$ two times we get
\begin{align}
F''(\alpha)&=\int_0^{\pi/2}e^{-\alpha\tan x}\tan^2x\,dx=-F(\alpha)+\int_0^{\pi/2}e^{-\alpha\tan x}(\underbrace{1+\tan^2x}_{\tan'(x)})\,dx=\\
&=-F(\alpha)+\int_0^{+\infty}e^{-\alpha t}\,dt=-F(\alpha)+\frac{1}{\alpha}.
\end{align}
The resulting differential equation
$$
F''+F=\frac{1}{\alpha},\quad\alpha>0,
$$
can be solved e.g. by variation of parameters
$$
F(\alpha)=A(\alpha)\cos\alpha+B(\alpha)\sin\alpha
$$
that gives
$$
\begin{bmatrix}
\cos\alpha & \sin\alpha\\
-\sin\alpha & \cos\alpha
\end{bmatrix}
\begin{bmatrix}
A'\\B'
\end{bmatrix}=\begin{bmatrix}
0\\1/\alpha
\end{bmatrix}\quad\Rightarrow\quad
\begin{bmatrix}
A'\\B'
\end{bmatrix}=\begin{bmatrix}
-\frac{\sin\alpha}{\alpha}\\\frac{\cos\alpha}{\alpha}
\end{bmatrix}.
$$
Together with the initial conditions
$$
F(+\infty)=F'(+\infty)=0
$$
it gives the solution
$$
F(\alpha)=-\cos\alpha\int_{+\infty}^\alpha\frac{\sin t}{t}\,dt+\sin\alpha\int_{+\infty}^\alpha\frac{\cos t}{t}\,dt.
$$
Now $\alpha=\frac{\pi}{2}$ gives the result.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
\int_{0}^{\pi/2}\expo{-\pi\tan\pars{t}/2}\,\dd t &
\,\,\,\stackrel{x\ =\ \tan\pars{t}}{=}\,\,\,
\int_{0}^{\infty}{\expo{-\pi x/2} \over x^{2} + 1}\,\dd x =
\Im\int_{0}^{\infty}{\expo{-\pi x/2} \over x - \ic}\,\dd x
\\[5mm] & =
\Im\int_{-\ic}^{\infty - \ic}{\expo{-\pi\ic/2}\expo{-\pi x/2} \over x}\,\dd x =
-\,\Re\int_{-\ic}^{\infty - \ic}{\expo{-\pi x/2} \over x}\,\dd x
\\[5mm] & \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}
\Re\int_{\infty}^{\epsilon}{\expo{-\pi x/2} \over x}\,\dd x +
\Re\int_{0}^{-\pi/2}
{\epsilon\expo{\ic\theta}\ic\,\dd\theta \over \epsilon\expo{\ic\theta}} +
\Re\int_{-\epsilon}^{-1}{\expo{-\pi\ic y/2} \over \ic y}\,\ic\,\dd y
\\[5mm] & \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}
-\int_{\pi\epsilon/2}^{\infty}{\expo{-x} \over x}\,\dd x -
\int^{\pi\epsilon/2}_{1}{\cos\pars{x} \over x}\,\dd y
\\[5mm] & \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\to}\,\,\,
\int^{\infty}_{\pi\epsilon/2}{\cos\pars{x} \over x}\,\dd y -
\int_{\pi\epsilon/2}^{\infty}{\expo{-x} \over x}\,\dd x -
\int^{\infty}_{\pi/2}{\cos\pars{x} \over x}\,\dd y
\\[5mm] & =
-\,\mrm{Ci}\pars{{\pi \over 2}\,\epsilon} -
\,\mrm{Ei}\pars{{\pi \over 2}\,\epsilon} + \,\mrm{Ci}\pars{\pi \over 2}
\label{1}\tag{1}
\end{align}

$\ds{\mrm{Ei}}$ is the
  Exponential Integral Function.
  Note that, as $\ds{z \to 0}$,
  $\ds{\,\mrm{Ci}\pars{z} \sim \gamma + \ln\pars{z} + \,\mrm{O}\pars{z^{2}}}$ and
  $\ds{\,\mrm{Ei}\pars{z} \sim -\gamma - \ln\pars{z} + \,\mrm{O}\pars{z^{1}}}$ 

such that \eqref{1} becomes
$$
\bbx{\int_{0}^{\pi/2}\expo{-\pi\tan\pars{t}/2}\,\dd t =
\,\mrm{Ci}\pars{\pi \over 2}}
$$
