For some reason I couldn’t find an answer to this online even though it seems very basic. I am trying to evaluate the following improper integral.

$ \int_0^{\infty} \frac{ \sin{x}}{x^2+1}dx$.

I consider the following function on nonnegative reals. $I(a) = \int_0^{\infty} \frac{ \sin{ax}}{x^2+1}dx$.

Differentiating under the integral sign I get $I'(a) = a\int_0^{\infty} \frac{ \cos{ax}}{x^2+1}dx$.

The rhs I evaluate with a contour integral to get $I'(a) = \frac{\pi}{2}ae^{-a}$.

With the boundary condition $I(0)=0$, I get $I(a) = \frac{\pi}{2}(1-e^{-a} -ae^{-a} )$. $a=1$ corresponds to the integral I want to evaluate, which is equal to

$$\frac{\pi}{2} - \frac{\pi}{e}$$

This doesn’t quite match the answer I am getting on [Wolfram Alpha][1]. Can someone please help me find out where I am going wrong? Thanks.


2 Answers 2


This question is answered in some of the questions listed in comments. However, I've given a slightly different approach to a slightly generalized question. This question is the case $a=1$ covered by this answer. $$ \begin{align}\newcommand{\Ei}{\operatorname{Ei}}\newcommand{\PV}{\operatorname{PV}} \int_0^\infty\frac{a\sin(x)}{a^2+x^2}\,\mathrm{d}x &=\frac1{2i}\int_0^\infty\frac{ae^{ix}-ae^{-ix}}{a^2+x^2}\,\mathrm{d}x\tag{1a}\\ &=\frac12\int_0^{-i\infty}\frac{ae^{-x}}{a^2-x^2}\,\mathrm{d}x+\frac{a}2\int_0^{i\infty}\frac{ae^{-x}}{a^2-x^2}\,\mathrm{d}x\tag{1b}\\ &=\PV\int_0^\infty\frac{ae^{-x}}{a^2-x^2}\,\mathrm{d}x\tag{1c}\\ &=\frac12\PV\int_0^\infty\frac{e^{-x}}{a+x}\mathrm{d}x+\frac12\PV\int_0^\infty\frac{e^{-x}}{a-x}\mathrm{d}x\tag{1d}\\ &=\frac{e^a}2\PV\int_a^\infty\frac{e^{-x}}{x}\mathrm{d}x-\frac{e^{-a}}2\PV\int_{-a}^\infty\frac{e^{-x}}{x}\mathrm{d}x\tag{1e}\\ &=-\frac{e^a}2\Ei(-a)+\frac{e^{-a}}2\Ei(a)\tag{1f} \end{align} $$ $\text{(1a):}$ $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$
$\text{(1b):}$ substitute $x\mapsto ix$ on the left and $x\mapsto-ix$ on the right
$\text{(1c):}$ bring $[0,i\infty]$ to the top of the positive real axis, and $[0,-i\infty]$ to the bottom
$\phantom{\text{(1c):}}$ being in opposite orientations, the bumps around $z=a$ cancel, leaving the
$\phantom{\text{(1c):}}$ principal value
$\text{(1d):}$ partial fractions
$\text{(1e):}$ substitute $x\mapsto x-a$ on the left and $x\mapsto x+a$ on the right
$\text{(1f):}$ apply the special function $\Ei(z)=-\PV\int_{-z}^\infty\frac{e^{-t}}{t}\,\mathrm{d}t$

Explanation of $\pmb{(1c)}$

enter image description here

Moving from $\text{(1b)}$ to $\text{(1c)}$ crosses no singularities and the integrals along the dashed curves vanish as the radius increases. The solid green arc contributes $-\pi i$ times the residue at $a$, while the solid red arc contributes $\pi i$ times the residue at $a$. That is, they cancel each other, leaving twice the Principal Value.


$\def\Ci{\operatorname{Ci}}\def\Si{\operatorname{Si}} \def\Chi{\operatorname{Chi}}\def\Shi{\operatorname{Shi}}$

Alternatively, we use $\Ci(z),\Si(z),\Chi(z),\Shi(z)$ functions:

$$\int_0^\infty \frac{\sin(x)}{x^2+1}dx=\frac i2\int_0^\infty\frac{\sin(x)}{x+i}dx-\frac i2\int_0^\infty \frac{\sin(x)}{x-i}dx= \frac i2\int_i^{i+\infty}\frac{\sin(x-i)}xdx-\frac i2\int_{-i}^{\infty-i} \frac{\sin(x+i)}xdx $$

Then expand $\sin(x\pm i)$:

$$\frac12\int_i^{i+\infty} \frac{\sinh(1)\cos(x)}{x}+\frac{i\cosh(1)\sin(x)}xdx+\frac12\int_{-i}^{\infty-i}\frac{\sinh(1)\cos(x)}x-\frac{i\cosh(1)\sin(x)}xdx$$

Plugging in gives:

$$\frac12(\sinh(1)\Ci(x)+i\cosh(1)\Si(x))|_i^{\infty+i}+\frac12(\sinh(1)\Ci(x)-i\cosh(1)\Si(x))|_{-i}^{\infty-i}=\frac12(\Shi(1)\cosh(1)-\Ci(-i)\sinh(1))+\frac12(\Shi(1)\cosh(1)-\Ci(i)\sinh(1))=\boxed{\Shi(1)\cosh(1)-\Chi(1)\sinh(1)= 0.64676112277913\dots}$$


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