How to evaluate $\int\limits_1^\sqrt3\frac{{e^{-x}}{sinx}}{x^2+1}dx$? I've tried integration by parts but it is not converging and the terms keep on increasing with every time I integrate by parts. Also, I can't guess any substitution and don't know if there is a property of definite integrals to be used here.
So, how would you solve it?
 A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\color{#f00}{\int_{1}^{\root{3}}{\expo{-x}\sin\pars{x} \over x^{2} + 1}\,\dd x} =
\Im\int_{1}^{\root{3}}{\expo{-x}\sin\pars{x} \over x - \ic}\,\dd x =
\Im\int_{1}^{\root{3}}{\expo{-x}\pars{\expo{\ic x} - \expo{-\ic x}}/\pars{2\ic} \over x - \ic}\,\dd x
\\[5mm] = &\
-\,{1 \over 2}\,\Re\int_{1}^{\root{3}}{\expo{\pars{-1 + \ic}x} \over x - \ic}
\,\dd x +
{1 \over 2}\,\Re\int_{1}^{\root{3}}{\expo{\pars{-1 - \ic}x} \over x - \ic}
\,\dd x
\end{align}
Now, you are close to the Exponential Integral. Can you take it from here ?.
A: It is not simple but doable using special functions.
Consider $$I=\int \frac{e^{-x} \sin (x)}{x^2+1}dx\qquad J=\int \frac{e^{-x} \cos (x)}{x^2+1}dx$$$$K=J+iI=\int \frac{e^{-x} e^{ix}}{x^2+1}dx=\int \frac{e^{-(1+i)x}} {x^2+1}dx$$ Now, use partial fraction decomposition $$\frac 1{x^2+1}=\frac{i}{2 (x+i)}-\frac{i}{2 (x-i)}$$ So $$K=\frac i 2\int \frac{e^{-(1+i)x}} {x+i}dx-\frac i 2\int \frac{e^{-(1+i)x}} {x-i}dx$$ For the first integral, change variable $x+i=y$ and for the second $x-i=z$.
The problem is that all of this will lead to exponential integrals of complex arguments $$K=\frac{1}{2} i e^{-1+i} \text{Ei}((1-i)-(1+i) x)-\frac{1}{2} i e^{1-i}
   \text{Ei}((-1-i) x-(1-i))$$
