Prove that: $ \int_{0}^{\infty} \frac{2 x \sin x+\cos 2x-1}{2 x^2} = 0$ How would you prove that?
$$ \int_{0}^{\infty} \frac{2 x \sin x+\cos 2x-1}{2 x^2} dx= 0$$
I'm looking for a solution at high school level if possible. Thanks.
 A: $\cos 2x=1-2\sin^2x$ 
so, $$\frac{2x\sin x+\cos 2x-1}{2x^2}=\frac{2x\sin x-2\sin^2 x}{2x^2}=\frac{\sin x}{x}-\frac{\sin^2 x}{x^2}$$
so, $$\int_0^{\infty}\frac{2x\sin x+\cos 2x-1}{2x^2}dx=\int_0^{\infty}\frac{\sin x}{x}dx-\int_0^{\infty}\frac{\sin^2 x}{x^2}dx$$
Now, $$\int_0^{\infty}\frac{\sin x}{x}dx=\pi/2$$
and $$\int_0^{\infty}\frac{\sin^2 x}{x^2}dx=\frac{-\sin^2 x}{x}|_0^{\infty}+\int_0^{\infty}\frac{2\sin x\cos x}{x}dx=0+\int_0^{\infty}\frac{\sin 2x}{2x}d(2x)=\pi/2$$
Thus, $$\int_0^{\infty}\frac{2x\sin x+\cos 2x-1}{2x^2}dx=\pi/2-\pi/2=0$$
For proof of $\int_0^{\infty}\frac{\sin x}{x}dx=\pi/2$
visit Evaluating the integral $\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$?
A: I have an answer for your old question without using 
$$\int_0^{\infty}\frac{\sin x}{x}dx=\frac\pi 2.$$
Noting that
$$ \int_0^{\infty}te^{-xt}dt=\frac{1}{x^2} $$
we have
\begin{eqnarray}
I&=&\int_{0}^{\infty} \frac{2 x \sin x+\cos 2x-1}{2 x^2} dx\\
&=& \frac{1}{2}\int_{0}^{\infty}(2 x \sin x+\cos 2x-1)\left(\int_0^{\infty}te^{-xt}dt\right) dx\\
&=&\frac{1}{2}\int_{0}^{\infty}\int_0^{\infty}(2 x \sin x+\cos 2x-1)te^{-xt}dt dx\\
&=&\frac{1}{2}\int_{0}^{\infty}t\left(\int_0^{\infty}(2 x \sin x+\cos 2x-1)e^{-xt}dx\right) dt\\
&=&2\int_{0}^{\infty}t\left(\frac{t}{(1+t^2)^2}-\frac{1}{4t+t^3}\right) dt\\
&=&2\int_{0}^{\infty}\left(\frac{t^2}{(1+t^2)^2}-\frac{1}{4+t^2}\right) dt.
\end{eqnarray}
It is very easy to calculate that
$$ \int_{0}^{\infty}\frac{t^2}{(1+t^2)^2}dt=\int_{0}^{\infty}\frac{1}{4+t^2}dt=\frac{\pi}{4} $$
so that 
$$I=0.$$
Done.
