Wolfram Alpha Can't Do This, Can You?

Despite the fact that I've dealed with this integral before yet it took more than an hour for me to solve it. I believe you're better than me. This is my approach:

First, we manipulate the trigonometric function. $$ $\begin{aligned} \int_0^\infty\frac{e^{-x\sqrt{3}}}{x}\left(1-\sin x\right)\left(1+2\sin x-\cos 2x\right)\;dx&=\int_0^\infty\frac{e^{-x\sqrt{3}}}{x}\left(1-\sin x\right)\cdot2\left(\sin x+\frac{1-\cos 2x}{2}\right)\;dx\\ &=2\int_0^\infty\frac{e^{-x\sqrt{3}}}{x}\left(1-\sin x\right)\left(\sin x+\sin^2 x\right)\;dx\\ &=2\int_0^\infty\frac{e^{-x\sqrt{3}}}{x}\left(1-\sin x\right)\cdot\sin x\left(1+\sin x\right)\;dx\\ &=2\int_0^\infty\frac{e^{-x\sqrt{3}}}{x}\sin x\left(1-\sin^2 x\right)\;dx\\ &=2\int_0^\infty\frac{e^{-x\sqrt{3}}}{x}\sin x\cos^2 x\;dx\\ &=\int_0^\infty\frac{e^{-x\sqrt{3}}}{x}(2\sin x\cos x)\cos x\;dx\\ &=\int_0^\infty\frac{e^{-x\sqrt{3}}}{x}\sin 2x\cos x\;dx\\ &=\int_0^\infty\frac{e^{-x\sqrt{3}}}{x}\cdot\frac{1}{2}(\sin(2x+x)+\sin(2x-x))\;dx\\ &=\frac{1}{2}\int_0^\infty\frac{e^{-x\sqrt{3}}}{x}(\sin 3x+\sin x)\;dx\\ &=\frac{1}{2}\left(\int_0^\infty\frac{e^{-x\sqrt{3}}}{x}\sin 3x\;dx+\int_0^\infty\frac{e^{-x\sqrt{3}}}{x}\sin x\;dx\right) \end{aligned}$ $$ From this point, it should be easier. You can either use double integral method like my solution in problem Definite Integral shared by Pranav Arora or you can use Integral Laplace transform . I prefer the second one because it's faster and simpler.

The Laplace transform of a function $f(x)$ with $x \ge0$ is given by $$ $\mathcal{L}[f(x)]=F(p)=\int_0^\infty e^{-px}f(x)\;dx.$ $$ Based on the table of Laplace transforms for $$ $f(x)=\frac{\sin ax}{x},$ $$ its Laplace transform is $$ $\mathcal{L}\left[\frac{\sin ax}{x}\right]=\tan^{-1}\left(\frac{a}{p}\right).$ $$ Therefore $$ $\begin{aligned} \frac{1}{2}\left(\int_0^\infty e^{-x\sqrt{3}}\left(\frac{\sin 3x}{x}\right)\;dx+\int_0^\infty e^{-x\sqrt{3}}\left(\frac{\sin x}{x}\right)\;dx\right)&=\frac{1}{2}\left(\tan^{-1}\left(\frac{3}{\sqrt{3}}\right)+\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)\right)\\ &=\frac{1}{2}\left(\frac{\pi}{3}+\frac{\pi}{6}\right)\\ &=\frac{\pi}{4}\\ &\approx\boxed{\color{#3D99F6}{0.7854}} \end{aligned}$ $$

$\text{\# }\mathbb{Q.E.D.}\text{ \#}$

Mathematica /can/ do this integral exactly, it just needs quite a bit of time to do so.

WolframAlpha will be happy to do this numerically. Substitute 0 in the prompt with 0.0 and it quickly returns the numerical approximation.

WolframAlpha Prompt

Yes, i did it , just like tunk :)

Manipulate the integrand as exp(-x√3)cos(x){sin(2x)/x} and integrate from {0 to infinity} in wolfram alpha to get answer=(π/4)=~0.785