To find closed form of $\int_0^{\frac{\pi}{2}} e^{-x\tan t+\alpha t} \;dt $ Let $x\geq 0$, then
$$\int_0^{\frac{\pi}{2}} e^{-x\tan t+\alpha t} \;dt = U_{\alpha} (x) $$
$$-\int_0^{\frac{\pi}{2}} \tan t \ e^{-x\tan t+\alpha t} \;dt = \frac{d (U_{\alpha} (x) )}{dx} $$
$$\int_0^{\frac{\pi}{2}} \tan^2 t \ e^{-x\tan t+\alpha t} \;dt = \frac{d^2(U_{\alpha} (x) )}{dx^2} $$
$$\int_0^{\frac{\pi}{2}} [-x(1+\tan^2 t)+\alpha] \ e^{-x\tan t+\alpha t} \;dt = -x\frac{d^2(U_{\alpha} (x) )}{dx^2}+ (\alpha-x) U_{\alpha} (x)$$
$$ \ e^{-x\tan t+\alpha t} |_0^{\frac{\pi}{2}} = -x\frac{d^2(U_{\alpha} (x) )}{dx^2}+ (\alpha-x) U_{\alpha} (x)$$
$$  -x\frac{d^2(U_{\alpha} (x) )}{dx^2}+ (\alpha-x) U_{\alpha} (x)=-1$$
$$  x\frac{d^2(U_{\alpha} (x) )}{dx^2}+ (x-\alpha) U_{\alpha} (x)=1$$
Boundry conditions:  for x=0 
$$\int_0^{\frac{\pi}{2}} e^{\alpha t} \;dt = U_{\alpha} (0) $$
$$\int_0^{\frac{\pi}{2}} e^{\alpha t} \;dt = U_{\alpha} (0)= \frac{e^{\frac{\pi}{2}\alpha}-1}{\alpha}$$
for $x=\infty $  ,$U_{\alpha} (\infty)=0$
Could you please help me to find the closed form of $U_{\alpha} (x)$?
Series and special function expressions all are welcome.
Thanks a lot for answers
 A: I really like this question and am curious to know the context in which such a function arises.
Denote $\delta_x = x \frac{\mathrm{d}}{\mathrm{d} x}$, and let's rewrite your differential equation with it: $\delta_x y(x) = x y^\prime(x)$ and $\delta_x^2 y(x) = x^2 y^{\prime\prime}(x) + x y^\prime(x)$, giving
$$
      \delta_x \left( \delta_x - 1\right) y(x) + x(x-\alpha) y(x) = x
$$
Apply the annihilator of the right handside $\delta_x - 1$ to both sides yields:
$$
    \left( \delta_x \left( \delta_x - 1 \right)^2 + x^2 \left(\delta_x + 1\right)  - \alpha x \delta_x \right) y(x) = 0 \tag{1}
$$
The differential equation is of a hypergeometric type only if $\alpha = 0$, in which case the solution can be read off in terms of Meijer's G-functions:
$$
     y_{\alpha=0}(x) = c_1 \cdot \cos(x) + c_2 \cdot \sin(x) + c_3 \cdot G_{1,3}^{3,1}\left(\frac{x^2}{4}\left|
\begin{array}{l}
 \frac{1}{2} \\
 \frac{1}{2},\frac{1}{2},0 \\
\end{array}\right.
\right)
$$
Because the integral tends to zero for large $x$, $c_1=c_2=0$. The integral, for $\alpha=0$ succumbs to the Mellin convolution technique:
$$
    \int_0^{\pi/2} \mathrm{e}^{-x \tan(t)}\mathrm{d} t = \int_0^\infty \mathrm{e}^{-u x} \frac{1}{1+u^2}\mathrm{d} u = \int_0^\infty f_1(x u) f_2\left(\frac{1}{u}\right) \frac{\mathrm{d} u}{u}
$$
where $f_1(u) = \mathrm{e}^{-u} = G_{0,1}^{1,0}\left(u\left|
\begin{array}{c}
 0 \\
\end{array}
\right.\right)$ and $f_2(u) = \frac{u}{1+u^2} = G_{1,1}^{1,1}\left(u^2\left|
\begin{array}{c}
 \frac{1}{2} \\
 \frac{1}{2} \\
\end{array} \right.
\right)$, thus getting:
$$
   \int_0^{\pi/2} \mathrm{e}^{-x \tan(t)}\mathrm{d} t = \frac{1}{2 \sqrt{\pi}} G_{1,3}^{3,1}\left(\frac{x^2}{4}\left|
\begin{array}{l}
 \frac{1}{2} \\
 \frac{1}{2},\frac{1}{2},0 \\
\end{array}\right.
\right) = \mathrm{Ci}(x) \sin (x)+\left(\frac{\pi }{2}-\mathrm{Si}(x)\right) \cos (x)
$$
where $\mathrm{Ci}(x) = - \int_{x}^\infty \frac{\cos(t)}{t} \mathrm{d} t$ and $\mathrm{Si}(x) = \int_0^x \frac{\sin(t)}{t} \mathrm{d} t$, i.e. $c_3 = \frac{1}{2 \sqrt{\pi}}$.
Notice that large $x$ asymptotic of the $\alpha=0$ solution is $y_{\alpha=0}(x) = \frac{1}{x} - \frac{2}{x^3} + \mathcal{o}\left( x^{-3} \right)$, i.e. it decreases algebraically, rather than exponentially.
Now return to equation $(1)$. Since the solution we seek decays exponentially for large $x$, its Mellin transform converges for $\Re(s)>0$:
$$
    \hat{y}_\alpha(s) = \int_0^\infty x^{s-1} y_\alpha(x) \mathrm{d} x 
$$
The differential equation $(1)$ for $y_\alpha(x)$ translates into a recurrence equation for $\hat{y}_\alpha(s)$:
$$
   (s+1) \left( s (s+1) \hat{y}_\alpha(s) + \hat{y}_\alpha(s) - \alpha  \, \hat{y}_\alpha(s+1) \right) = 0 \tag{2}
$$
Rather than solve such an equation directly, it is easier to find $\hat{y}_\alpha(s)$ directly:
$$ \begin{eqnarray}
  \hat{y}_\alpha(s) &=& \int_0^\infty x^{s-1} y_\alpha(x) \mathrm{d} x = \int_0^{\pi/2} \mathrm{e}^{\alpha t} \int_0^{\infty} x^{s-1} \mathrm{e}^{-x \tan(t)} \mathrm{d} x \mathrm{d} t = \Gamma(s) \int_0^{\pi/2} \mathrm{e}^{\alpha t} \tan^{-s}(t) \mathrm{d} t \\
  &=& \Gamma(s) \int_0^\infty \mathrm{e}^{\alpha \arctan(u)} \frac{u^{-s}}{1+u^2}  \mathrm{d} u = \Gamma(s) \int_0^\infty u^{-s} \left( 1 + i u \right)^{i \frac{\alpha}{2} -1} \left( 1 - i u \right)^{-i \frac{\alpha}{2} -1} \mathrm{d} u \\
   &=& \frac{\Gamma \left(1-\frac{i \alpha }{2}\right) e^{\frac{1}{2} \pi  (\alpha -i s)} \Gamma (s) \Gamma (s+1) }{\alpha  \Gamma
   \left(s-\frac{i \alpha }{2}+1\right)} {}_2F_1\left(s,-\frac{i \alpha }{2};s-\frac{i \alpha }{2}+1;-1\right) \\ &\phantom{=}&-\frac{\pi  e^{-\frac{1}{2} i \pi  s} \Gamma \left(1-\frac{i \alpha }{2}\right)  }{\alpha  \Gamma \left(-s-\frac{i \alpha }{2}+1\right) \sin(\pi s)} {}_2F_1\left(-s,-\frac{i \alpha }{2};-s-\frac{i \alpha
   }{2}+1;-1\right)
\end{eqnarray} \tag(3)
$$
where $0 < \Re(s) < 1$ is needed for convergence, in agreement with the algebraic tail of $\alpha=0$ case. Thus the solution admits Mellin-Barnes integral representation:
$$
    y_\alpha(x) = \frac{1}{2 \pi i}\int_{\gamma - i \infty }^{\gamma+i \infty} \hat{y}_\alpha(s) x^{-s} \mathrm{d} s
$$
where $\gamma$ is arbitrary real, such that $0<\gamma<1$.
It is not hard to verify that equation $(3)$ verifies the recurrence relation $(2)$.

Series at the origin (added)
The differential equation determines the series expansion for the fundamental system, but the Mellin-Barnes representation allows you to find the series as a sum of residue. The sum of residues at $s=0$ and $s=1$ gives the following series: 
$$y_\alpha(z) = \frac{1}{\alpha} \left(\mathrm{e}^{\frac{\pi}{2} \alpha}-1 \right) + \mathrm{e}^{\frac{\pi}{2} \alpha} z \log(z) + \frac{z}{2\alpha} \left( -\frac{4}{\alpha+2i} {}_2F_1(1,-i \frac{\alpha}{2} ; 2-i \frac{\alpha}{2}; -1 ) + \mathrm{e}^{\frac{\pi \alpha}{2}} \left(2 i + ( 4 \gamma + i \pi -1 ) \alpha + 2 \alpha \psi\left(- i \frac{\alpha}{2} \right)  \right)  \right) + \mathcal{o}(z)$$ 


Series at $\infty$ (added)
Having the Mellin-Barnes representation handy allows to also determine the series expansion for large $x$. One would have $y_\alpha(x) = -\sum_{k=1}^{n} \operatorname{Res}_{s=k} \hat{y}_\alpha(s) x^{-s} + \mathcal{o}\left(x^{-n}\right)$. Computing residues gives:
$$
    y_\alpha(x) = \frac{1}{x} + \frac{\alpha}{x^2} + \frac{\alpha^2-2}{x^3} + \frac{\alpha^3-8 \alpha}{x^4} + \mathcal{o}\left(x^{-4}\right)
$$

The expansion can be easily verified to agree with the differential equation.
A: For $\alpha=0$, Maple says this:
$$
\int_{0}^{\frac{\pi}{2}} \operatorname{e} ^{-x \operatorname{tan} t} d t = \frac{i \operatorname{e} ^{i x}}{2}  \mathrm{Ei}_1 (i x) - \frac{i \operatorname{e} ^{-i x}}{2} \mathrm{Ei}_1 (-ix)
$$
in terms of the exponential integral
$$
\mathrm{Ei}_a(z) = \int_1^\infty \operatorname{e}^{-sz} s^{-a}\,ds
$$
For the DE, Maple has an answer involving integrals and the Whittaker M and W functions.

_C1 and _C2 are two arbitrary constants.  It looks much worse than the original way of writing this!
