# 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?

• Add your working into your post so we can see what you have done. – Mattos Sep 27 '16 at 4:23
• What makes you think it has a closed form? – MathematicsStudent1122 Sep 27 '16 at 4:25
• Is the $e^{-x}$ inside or outside the $\sin$? I really hope it's outside... – Carl Schildkraut Sep 27 '16 at 4:30
• @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? – Carl Schildkraut Sep 27 '16 at 4:43
• @ankit No, because of the $+1$ in the denominator. – Carl Schildkraut Sep 27 '16 at 5:00

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 $$K=\frac{1}{2} i e^{-1+i} \text{Ei}((1-i)-(1+i) x)-\frac{1}{2} i e^{1-i} \text{Ei}((-1-i) x-(1-i))$$