# How to evaluate the following improper integral : \int_0^{+\infty}\frac{x\sin x}{x^2+1}dx

How to evaluate the following improper integral :
$$\displaystyle{\int_{{0}}^{{+\infty}}}{\frac{{{x}{\sin{{x}}}}}{{{x}^{{2}}+{1}}}}{\left.{d}{x}\right.}$$

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Navreaiw
According to the residue theory,
$$\displaystyle{\int_{{0}}^{{+\infty}}}{\frac{{{1}}}{{{s}^{{2}}+{x}^{{2}}}}}{\left.{d}{x}\right.}={\frac{{\pi}}{{{2}{s}}}},\ {I}{\left(\alpha\right)}={\int_{{0}}^{{+\infty}}}{\frac{{{x}{\sin{\alpha}}{x}}}{{{1}+{x}^{{2}}}}}{\left.{d}{x}\right.}$$
Laplace transform:
$$\displaystyle{L}{\left[{I}{\left(\alpha\right)}\right]}={\int_{{0}}^{{+\infty}}}{\frac{{{x}}}{{{1}+{x}^{{2}}}}}\cdot{\frac{{{x}}}{{{s}^{{2}}+{x}^{{2}}}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\int_{{0}}^{{+\infty}}}{\frac{{{x}^{{2}}+{1}-{1}}}{{{1}+{x}^{{2}}}}}\cdot{\frac{{{1}}}{{{s}^{{2}}+{x}^{{2}}}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\int_{{0}}^{{+\infty}}}{\frac{{{1}}}{{{s}^{{2}}+{x}^{{2}}}}}{\left.{d}{x}\right.}-{\int_{{0}}^{{+\infty}}}{\frac{{{1}}}{{{1}+{x}^{{2}}}}}\cdot{\frac{{{1}}}{{{s}^{{2}}+{x}^{{2}}}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\int_{{0}}^{{+\infty}}}{\frac{{{1}}}{{{s}^{{2}}+{x}^{{2}}}}}{\left.{d}{x}\right.}-{\frac{{{1}}}{{{s}^{{2}}-{1}}}}{\int_{{0}}^{{+\infty}}}{\left({\frac{{{1}}}{{{1}+{x}^{{2}}}}}-{\frac{{{1}}}{{{s}^{{2}}+{x}^{{2}}}}}\right)}{\left.{d}{x}\right.}$$
$$\displaystyle={\frac{{\pi}}{{{2}}}}\cdot{\frac{{{1}}}{{{s}+{1}}}}$$
Inverse transform:
$$\displaystyle{L}^{{-{1}}}{\left[{I}{\left(\alpha\right)}\right]}={\frac{{\pi}}{{{2}}}}{e}^{{-\alpha}}\Rightarrow{I}{\left({1}\right)}={\int_{{0}}^{{+\infty}}}{\frac{{{x}{\sin{{x}}}}}{{{1}+{x}^{{2}}}}}{d}={\frac{{\pi}}{{{2}{e}}}}$$
###### Not exactly what youâ€™re looking for?
Juan Spiller
Using the result from this OP: Integral evaluation $$\displaystyle{\int_{{-\infty}}^{{\infty}}}{\frac{{{\cos{{\left({a}{x}\right)}}}}}{{\pi{\left({1}+{x}^{{2}}\right)}}}}{\left.{d}{x}\right.}$$. We have
$$\displaystyle{\int_{{0}}^{{\infty}}}{\frac{{{\cos{{a}}}{x}}}{{{1}+{x}^{{2}}}}}{\left.{d}{x}\right.}={\frac{{\pi}}{{{2}}}}{e}^{{-{\left|{a}\right|}}}$$
Thus, our integration is simply
$$\displaystyle{\int_{{0}}^{\infty}}{\frac{{{x}{\sin{{x}}}}}{{{1}+{x}^{{2}}}}}{\left.{d}{x}\right.}=-\lim_{{{a}\to{1}}}{d}_{{a}}{\int_{{0}}^{{\infty}}}{\frac{{{\cos{{a}}}{x}}}{{{1}+{x}^{{2}}}}}{\left.{d}{x}\right.}$$
$$\displaystyle=-{\frac{{\pi}}{{{2}}}}\lim_{{{a}\to{1}}}{d}_{{a}}{\left[{e}^{{-{\left[{a}\right]}}}\right]}$$
$$\displaystyle={\frac{{\pi}}{{{2}{e}}}}$$
star233

Since your integrand is even, this integral is equal to one half of
$$\int_{-\infty}^\infty\frac{x\sin x}{x^2+1}dx$$
This integral is the imaginary part of
$$\int_{-\infty}^\infty\frac{x\cdot e^{ix}}{x^2+1}dx$$
This can be solved as the contour integral with contour a half disc of radius R with base on the real axis, letting R go to infinity. The integral along the arc goes to 0 ( needs some showing ). The contour integral is equal to the residue at $$x=i$$, which is $$e^{-1}\pi i$$. So taking half the imaginary part, we get $$\frac{\pi}{2e}.$$