Integral

**Random Variable** · April 22nd 2010, 07:28 PM

Challenge Question:

$\int^{\infty}_{0} \frac{\cos x - \cos 3x}{x^{2}} \ dx$

Moderator editor: Approved Challenge question.

**simplependulum** · April 22nd 2010, 09:19 PM

Originally Posted by Random Variable

Challenge Question:

$\int^{\infty}_{0} \frac{\cos x - \cos 3x}{x^{2}} \ dx$

Moderator editor: Approved Challenge question.

$\int_0^{\infty} \frac{ \cos x - \cos{3x} }{ x^2}~dx$

$= \int_0^{\infty} \int_1^3 \frac{\sin{xy}}{x}~dxdy$

$= \frac{\pi}{2} \int_1^3 dy = \pi$

**Drexel28** · April 22nd 2010, 09:31 PM

Originally Posted by Random Variable

Challenge Question:

$\int^{\infty}_{0} \frac{\cos x - \cos 3x}{x^{2}} \ dx$

Moderator editor: Approved Challenge question.

Alternatively. Let $J(\theta)=\int_0^{\infty}\frac{\cos(x)-\cos(\theta x)}{x^2}dx$ . Then, $J'(\theta)=\int_0^{\infty}\frac{\partial}{\partial \theta}\frac{\cos(x)-\cos(\theta x)}{x^2}dx=\int_0^{\infty}\frac{\sin(\theta x)}{x}dx$ . Let $\theta x=z$ to get $\int_0^{\infty}\frac{\sin(z)}{z}dz$ but this is a routine calculation ( $\frac{1}{z}=\int_0^{\infty}e^{-zt}dt$ ) and it equals $\frac{\pi}{2}$ . So, $J'(\theta)=\frac{\pi}{2}$ and so $J(\theta)=\frac{\pi\theta}{2}+C$ . But, $0=J(1)=\frac{\pi}{2}+C\implies C=-\frac{\pi}{2}$ . Thus, $J(\theta)=\frac{\pi}{2}\left(\theta-1\right)$

**Random Variable** · April 22nd 2010, 09:45 PM

or

$\int^{\infty}_{0} \frac{\cos x - \cos 3x}{x^{2}} \ dx = \int^{\infty}_{0} \int^{\infty}_{0} te^{-xt} (\cos x - \cos 3x) \ dt \ dx$

$= \int^{\infty}_{0} t \Big(\mathcal{L} \{\cos x\} - \mathcal{L} \{\cos 3x\} \Big) dt = \int^{\infty}_{0} \Big(\frac{t^{2}}{t^{2}+1} - \frac{t^{2}}{t^{2}+9} \Big) \ dt$

$= \int^{\infty}_{0} \Big(\frac{9}{t^{2}+9} - \frac{1}{1+t^{2}} \Big) \ dt = \Big(3 \tan (t/3) -\tan t\Big) \Big|^{\infty}_{0} = -\frac{\pi}{2} + \frac{3 \pi}{2} = \pi$

Thread: Integral

LinkBack

Thread Tools

Display

Integral

Similar Math Help Forum Discussions

Use Cauchy's integral theorem for derivatives to evaluate the contour integral.

Graph the region of integration and evaluate the double integral (integral from 0 to

Explaining result: comparing line integral and two-form integral

[SOLVED] Line integral, Cauchy's integral formula

write [integral of] dx/SQRT(sinx) in terms of an elliptic integral

Search Tags