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.

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    $\begingroup$ How about using the Taylor series? $\endgroup$ – The Chaz 2.0 Sep 5 '12 at 11:24
  • $\begingroup$ @The Chaz: a good idea $\endgroup$ – user 1357113 Sep 5 '12 at 11:25
  • $\begingroup$ Maybe there is a way that doesn't use the fact that $\int_0^{\infty}\frac{\sin x}{x}dx=\pi/2$. $\endgroup$ – user 1357113 Sep 5 '12 at 12:39

$\cos 2x=1-2\sin^2x$

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 $\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$?

  • $\begingroup$ This answer requires knowing that $\displaystyle \int_0^\infty \frac{\sin x}{x}\,\mathrm dx = \frac{\pi}{2}$. Can this be shown by methods available to beginning students in calculus? The OP did say that a solution at high school level was wanted, if one was possible. $\endgroup$ – Dilip Sarwate Sep 5 '12 at 12:07
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    $\begingroup$ @DilipSarwate Actually Avatar's answer doesn't require to know that $\int_0^{+\infty}\frac{\sin x}{x}dx = \frac{\pi}{2}$. All it requires to know is that this quantity is defined and equal to some constant $K$, then you can apply the same reasoning and the $K$'s will simplify themselves. $\endgroup$ – S4M Sep 5 '12 at 13:00
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    $\begingroup$ "All it requires to know that this quantity is defined and equal to some constant $K, \ldots$" itself is something that needs just a little bit of extra thought, doesn't it? How can we be sure that the improper integral has a finite value? But yes, once we can be sure that $\int_0^\infty \frac{\sin x}{x}\,\mathrm dx$ is finite, its exact value is immaterial in Avatar's proof. $\endgroup$ – Dilip Sarwate Sep 5 '12 at 14:34

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.


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