Integral

**Random Variable** · May 4th 2010, 11:53 AM

Challenge Problem:

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

Moderator Approved - CB

**NonCommAlg** · May 4th 2010, 07:45 PM

Originally Posted by Random Variable

Challenge Problem:

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

Moderator Approved - CB

starting with by parts we have:

$\int_0^{\infty} \frac{\sin^3x}{x^2} \ dx = 3 \int_0^{\infty} \frac{\sin^2 x \cos x}{x} \ dx = 3 \int_0^{\infty} \int_0^{\infty} \sin^2 x \cos x \ e^{-yx} \ dy \ dx$

$= 3 \int_0^{\infty} \int_0^{\infty} \sin^2 x \cos x \ e^{-yx} \ dx \ dy = \frac{3}{4} \int_0^{\infty} \int_0^{\infty}[\cos x - \cos(3x)]e^{-yx} \ dx \ dy$

$=\frac{3}{4} \int_0^{\infty} \left( \frac{y}{y^2+1} - \frac{y}{y^2+9} \right) \ dy = \frac{3}{8} \left [ \ln \left( \frac{y^2+1}{y^2+9} \right) \right]_0^{\infty} =\frac{3}{4} \ln 3.$

**Random Variable** · May 4th 2010, 08:16 PM

Originally Posted by NonCommAlg

starting with by parts we have:

$\int_0^{\infty} \frac{\sin^3x}{x^2} \ dx = 3 \int_0^{\infty} \frac{\sin^2 x \cos x}{x} \ dx = 3 \int_0^{\infty} \int_0^{\infty} \sin^2 x \cos x \ e^{-yx} \ dy \ dx$

$= 3 \int_0^{\infty} \int_0^{\infty} \sin^2 x \cos x \ e^{-yx} \ dx \ dy = \frac{3}{4} \int_0^{\infty} \int_0^{\infty}[\cos x - \cos(3x)]e^{-yx} \ dx \ dy$

$=\frac{3}{4} \int_0^{\infty} \left( \frac{y}{y^2+1} - \frac{y}{y^2+9} \right) \ dy = \frac{3}{8} \left [ \ln \left( \frac{y^2+1}{y^2+9} \right) \right]_0^{\infty} =\frac{3}{4} \ln 3.$

Nice.

The following was my approach:

$\int^{\infty}_{0} \frac{\sin^{3} x}{x^{2}} \ dx = \frac{1}{4} \int^{\infty}_{0} \frac{3 \sin x - \sin 3x}{x^{2}} \ dx$ (triple-angle formula for $\sin$ )

$= \frac{1}{4} \lim_{\delta \to 0} \Big(3\int^{\infty}_{\delta} \frac{\sin x}{x^{2}} \ dx - \int^{\infty}_{\delta} \frac{\sin 3x} {x^{2}} \ dx \Big)$

$= \frac{1}{4} \lim_{\delta \to 0} \Big(3\int^{\infty}_{\delta} \frac{\sin x}{x^{2}} \ dx - \frac{1}{3}\int^{\infty}_{3 \delta} \frac{\sin u} {(\frac{u}{3})^{2}} \ du \Big)$

$= \frac{3}{4} \lim_{\delta \to 0} \Big(3\int^{\infty}_{\delta} \frac{\sin x}{x^{2}} \ dx - \int^{\infty}_{3 \delta} \frac{\sin u} {u^{2}} \ du \Big)$

$=\frac{3}{4} \lim_{\delta \to 0} \int_{\delta}^{3 \delta} \frac{\sin x}{x^{2}} \ dx$

$= \frac{3}{4} \lim_{\delta \to 0} \int_{\delta}^{3 \delta} \frac{1}{x} \ dx$ (since we're integrating over x-values close to 0)

$= \frac{3}{4} \lim_{\delta \to 0} (\ln {3\delta} - \ln {\delta}) = \frac{3}{4} \ln 3$

**pankaj** · June 29th 2010, 06:11 PM

Originally Posted by Random Variable

$=\frac{3}{4} \lim_{\delta \to 0} \int_{\delta}^{3 \delta} \frac{\sin x}{x^{2}} \ dx$

$= \frac{3}{4} \lim_{\delta \to 0} \int_{\delta}^{3 \delta} \frac{1}{x} \ dx$ (since we're integrating over x-values close to 0)

Is this justified since
$\lim_{x\to 0}\frac{\sin x}{x^2}$
does not exist.

Moreover,
$\lim_{x\to a}f(x).g(x)=\lim_{x\to a}f(x).\lim_{x\to a}g(x)$

only if both
$\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$
exist and are finite.

You may have obtained the answer but there are loopholes.

The solution is pretty good otherwise.

**simplependulum** · July 1st 2010, 03:18 AM

We can use the infinite series $\sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} - ....$

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