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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.

Aang
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user 1591719
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3 Answers3

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

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

Community
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Aang
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  • 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. – Dilip Sarwate Sep 05 '12 at 12:07
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    @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. – S4M Sep 05 '12 at 13:00
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    "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. – Dilip Sarwate Sep 05 '12 at 14:34
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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.

xpaul
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Using only double angle formula of $\cos 2x$ followed IBP, we have $$ \begin{aligned} \int_0^{\infty} \frac{2 x \sin x+\sin 2 x-1}{2 x^2} d x = & \int_0^{\infty} \frac{2 x \sin x-2 \sin ^2 x}{2 x^2} d x \\ = & \int_0^{\infty} \frac{\sin x}{x} d x+\int_0^{\infty} \sin ^2 x d\left(\frac{1}{x}\right) \\ = & \int_0^{\infty} \frac{\sin x}{x} d x+\left[\frac{\sin ^2 x}{x}\right]_0^{\infty}-\int_0^{\infty} \frac{\sin (2 x)}{x} d x \\ = & \int_0^{\infty} \frac{\sin x}{x} d x-\int_0^{\infty} \frac{\sin x}{x} d x \quad \textrm{ via } 2 x \mid \rightarrow x\\ = & 0 \end{aligned} $$

Lai
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