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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?

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    $\begingroup$ Add your working into your post so we can see what you have done. $\endgroup$ – Mattos Sep 27 '16 at 4:23
  • $\begingroup$ What makes you think it has a closed form? $\endgroup$ – MathematicsStudent1122 Sep 27 '16 at 4:25
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    $\begingroup$ Is the $e^{-x}$ inside or outside the $\sin$? I really hope it's outside... $\endgroup$ – Carl Schildkraut Sep 27 '16 at 4:30
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    $\begingroup$ @ankit The substitution $x=\tan\theta$ gives us $$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} e^{-\tan\theta}\sin\tan\theta\ d\theta$$ which might be a little nicer? $\endgroup$ – Carl Schildkraut Sep 27 '16 at 4:43
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    $\begingroup$ @ankit No, because of the $+1$ in the denominator. $\endgroup$ – Carl Schildkraut Sep 27 '16 at 5:00
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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 $$K=\frac{1}{2} i e^{-1+i} \text{Ei}((1-i)-(1+i) x)-\frac{1}{2} i e^{1-i} \text{Ei}((-1-i) x-(1-i))$$

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$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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 ?.

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