integral

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June 10th 2011, 03:06 PM
Random Variable

improper integral #3

Challenge Problem:

Show that $\displaystyle \int^{\infty}_{0} \frac{\cos (x^{2}) - \sin (x^{2})}{1+x^{4}} \ dx = \frac{\sqrt{2} \pi}{4e}$
June 10th 2011, 04:03 PM
DeMath

Can we use the residue theorem?
June 10th 2011, 04:10 PM
Random Variable

Quote:

Originally Posted by DeMath

Can we use the residue theorem?

You can use any method you want.
June 10th 2011, 10:48 PM
TheCoffeeMachine

I wonder whether the following fact is of any use here

$\int_{0}^{\infty}e^{-y}\cos(x^2y)\;{dy} = \frac{1}{1+x^4}$

I'm not familiar with double integrals, unfortunately! (Doh)
June 11th 2011, 11:52 AM
Random Variable

If you use that fact, then we have $\int^{\infty}_{0} \int^{\infty}_{0} \big(\cos(x^{2}) - \sin(x^{2}) \big) \cos (x^{2}y) e^{-y} \ dy \ dx$

But then changing the order of integration (which is justified because the original integral converges absolutely) doesn't really simply things.

The approach I had in mind was to use contour integration using a slightly unusual contour.
June 11th 2011, 01:27 PM
Random Variable

Spoiler:

Let $\displaystyle f(z) = \frac{e^{iz^{2}}}{1+z^{4}}$

And use the contour that consists of the line segment from the origin to (R,0), the arc from (R,0) to (0,iR), and the line segment back to the origin.

Interesting things happen.
June 11th 2011, 09:35 PM
chisigma

Quote:

Originally Posted by Random Variable

Challenge Problem:

Show that $\displaystyle \int^{\infty}_{0} \frac{\cos (x^{2}) - \sin (x^{2})}{1+x^{4}} \ dx = \frac{\sqrt{2} \pi}{4e}$

A method that uses the Laplace Transform: we define...

$\gamma(t)= \int_{0}^{\infty} \frac{\cos t x^{2} - \sin t x^{2}}{1+x^{4}}\ dx$ (1)

... so that is...

$\mathcal{L} \{\gamma(t)\} = \int_{0}^{\infty} e^{-s t}\ \int_{0}^{\infty} \frac{\cos t x^{2} - \sin t x^{2}}{1+x^{4}}\ dx\ dt =$

$= \int_{0}^{\infty} \frac{\mathcal{L} \{ \cos t x^{2} - \sin t x^{2} \}}{1+x^{4}}\ dx = \int_{0}^{\infty} \frac{s - x^{2}} {(1+x^{4})\ (s^{2}+x^{4})} \ dx$ (2)

To be continued in next posts...

Kind regards

$\chi$ $\sigma$
June 12th 2011, 07:52 AM
chisigma

Performing the partial fraction expansion od the result (2) of my previous post we obtain...

$\mathcal {L} \{\gamma(t)\} = \frac{s}{s^{2}-1}\ \{\int_{0}^{\infty} \frac{dx}{1+x^{4}} - \int_{0}^{\infty} \frac{dx}{s^{2}+x^{4}} \} -$

$- \frac{1}{s^{2}-1}\ \{\int_{0}^{\infty} \frac{x^{2}}{1+x^{4}}\ dx - \int_{0}^{\infty} \frac{x^{2}}{s^{2}+x^{4}}\ dx \}$ (3)

The integrals in x can be solved in standard way with an approriate path in the complex plane obtaining...

$\int_{0}^{\infty} \frac{dx}{1+x^{4}} = \frac{\pi}{2\ \sqrt{2}}$ (4)

$\int_{0}^{\infty} \frac{dx}{s^{2}+x^{4}} = \frac{\pi}{2\ \sqrt{2}\ s^{\frac{3}{2}}}$ (5)

$\int_{0}^{\infty} \frac{x^{2}}{1+x^{4}}\ dx = \frac{\pi}{2\ \sqrt{2}}$ (6)

$\int_{0}^{\infty} \frac{x^{2}}{s^{2}+x^{4}}\ dx = \frac{\pi}{2\ \sqrt{2}\ s^{\frac{1}{2}}}$ (7)

Now is we substitute (4),(5), (6) and (7) in (3) we obtain the [incredible] result...

$\mathcal {L} \{\gamma(t)\} = \frac{\pi}{2\ \sqrt{2}}\ \frac{1}{s+1}$ (8)

... so that is...

$\gamma (t) = \frac{\pi}{2\ \sqrt{2}}\ e^{-t}$ (9)

... and finally...

$\int_{0}^{\infty} \frac{\cos x^{2} - \sin x^{2}}{1+x^{4}}\ dx = \gamma (1) = \frac{\pi}{2\ e\ \sqrt{2}}$ (10)

Kind regards

$\chi$ $\sigma$
June 12th 2011, 10:10 AM
Random Variable

Let $f(z) = \frac{e^{iz^{2}}}{1+z^{4}}$

Let $C$ be the contour consisting of the line segment from the origin to (0,R); the arc from (R,0) to (0, IR); and the line segment from (0,IR) back to the origin.

$f(z)$ has 4 simple poles at $e^{\frac{i}{4}(\pi+2 \pi n)}, \ n=0,1,2,3$

The only pole inside of the contour is $e^{\frac{i \pi}{4}}$

So we have $\int_{C} f(z) \ dz = \int^{R}_{0} f(x) \ dx + \int_{arc} f(z) \ dz + \int^{0}_{R} f(it) \ i dt = 2 \pi i \ \text{Res} [f,e^{\frac{i \pi}{4}} ]$

$\text{Res}[f,e^{\frac{i \pi}{4}}] = \lim_{z \to e^{\frac{i \pi}{4}}} \ (z - e^{\frac{i \pi}{4}}) \frac{e^{iz^{2}}}{1+z^{4}} = \frac{e^{-\frac{3 i \pi}{4}}}{4 e}$ (using L'Hospital's rule)

$= \frac{1}{4e} \Big( \frac{- 1}{\sqrt{2}} - \frac{i}{\sqrt{2}} \Big)$

$\Big| \int_{arc} f(z) \ dz \Big| = \Big| \int^{\frac{\pi}{2}}_{0} \frac{e^{i(Re^{it})^{2}}}{1+ R^{4} e^{4it}} \ i Re^{it} \ dt \Big| \le \frac{\pi}{2} \frac{e^{-R^{2} \sin 2t}}{R^{4}-1} \ R$ (ML inequality and triangle inequality)

$\le \frac{\pi}{2} e^{-R^{2}} \frac{R}{R^{4}+1}$ (since $0 \le t \le \frac{\pi}{2}$ )

therefore $\lim_{R \to \infty} \Big| \int_{arc} f(z) \ dz \Big| \le 0$

$\int^{0}_{R} f(it) \ i dt = - \int^{R}_{0} \frac{e^{i(it)^{2}}}{1+ (it)^{4}} \ i dt = - i \int^{R}_{0} \frac{e^{-it^{2}}}{1+t^{4}} \ dt$

$= -i \int^{R}_{0} \frac{\cos (t^{2}) - i \sin (t^{2}) }{1+t^{4}} \ dt = -\int^{R}_{0} \frac{\sin (t^{2}) + i \cos (t^{2})}{1+t^{4}} \ dt$

Now we have $\int_{0}^{R} \frac{ \cos (x^{2}) + i\sin (x^{2})}{1+x^{2}} \ dx + \int_{arc} f(z) \ dz - \int_{0}^{R} \frac{ \sin (x^{2}) + i\cos (x^{2})}{1+x^{4}} \ dx$

$= 2 \pi i \Big( \frac{-1}{\sqrt{2}} - \frac{i}{\sqrt{2}} \Big) = \frac{\sqrt{2} \pi}{4e} (1-i)$

Now take the limit as R goes to infinity and equate the real portions on both sides to get

$= \int^{\infty}_{0} \frac{\cos (x^{2}) - \sin (x^{2})}{1+x^{4}} \ dx = \frac{\sqrt{2} \pi}{4e}$