Problem of the Week #2 [from Walking Randomly blog] – Submission

Posted on Julho 24, 2008 por

I’ve just submitted the following solution for the

« Integral of the Week #2

The second Integral Of The Week (IOTW) is rather different from the first in that I am going to give you the evaluation. Your task is to prove it.

\displaystyle\int_{-\infty}^{\infty}e^{-x^2}\; dx=\sqrt{\pi}

But WAIT! Almost every time I have seen this integral evaluated, it has been done by squaring it and converting to polar co-ordinates and that’s the one method of evaluation you can’t use for this particular challenge. I am looking for more ‘interesting’ proofs. Have fun.

Solutions can be posted in the comments section or sent to me by email (obtaining my email address is another puzzle for you to solve) and will be discussed in a future post. Feel free to send your solution in just about any format you like – plain text, uncompiled Latex, PDF, postscript, Mathematica, ODF, even Microsoft Word. When I get around to posting the solutions I will attempt to standardize them (to PDF probably).

By the way – you still have time to submit a solution for the first IOTW.

July 22nd, 2008 »

Solution

Firstly I show that

\displaystyle\int_{0}^{\infty}xe^{-x^{2}y^{2}}\; dy=I,

where

I=\displaystyle\int_{0}^{\infty}e^{-x^2}\; dx.

Then I evaluate I^2 to derive the value of I by reversing the order of integration.

The second Integral Of The Week equals 2I:

\displaystyle\int_{-\infty}^{\infty}e^{-x^2}\; dx=2I.

If x>0, then

\displaystyle\int_{0}^{\infty}xe^{-x^{2}y^{2}}\; dy=x\displaystyle\int_{0}^{\infty}e^{-{(xy)}^2}\; dy=x\displaystyle\int_{0}^{\infty}e^{-u^2}\dfrac{du}{x}=\displaystyle\int_{0}^{\infty}e^{-u^2}\; du=I.

Thus

I^2=\displaystyle\int_{0}^{\infty}e^{-x^2}\; dx\displaystyle\int_{0}^{\infty}xe^{-x^{2}y^{2}}\; dy =\displaystyle\int_{0}^{\infty}dy\displaystyle\int_{0}^{\infty}xe^{-x^2}e^{-x^{2}y^{2}}\; dx =\displaystyle\int_{0}^{\infty}dy\displaystyle\int_{0}^{\infty}xe^{-x^{2}(1+y^2)}\; dx =\displaystyle\int_{0}^{\infty}dy\dfrac{1}{2\left( 1+y^{2}\right) }\left[ -e^{-x^{2}\left( 1+y^{2}\right) }\right] _{x=0}^{\infty } =\displaystyle\int_{0}^{\infty }\dfrac{-e^{-\underset{x\rightarrow \infty}{\lim }x}+e^{0}}{2\left( 1+y^{2}\right) }\; dy =\displaystyle\int_{0}^{\infty }\dfrac{1}{2\left( 1+y^{2}\right) }\; dy

=\dfrac{1}{2}\left[ \tan^{-1}y\right] _{y=0}^{\infty } =\dfrac{1}{2}\left( \dfrac{\pi}{2}-0\right) =\dfrac{\pi}{4}.

So

I=\dfrac{\sqrt{\pi}}{2}

and

\displaystyle\int_{-\infty}^{\infty}e^{-x^2}\; dx=2I=\sqrt{\pi}.

\square

[Posted in my auxiliary blog Cálculos Auxiliares


http://calcauxprobteor.wordpress.com/2008/07/24/integral-of-the-week-2-walking-randomly/
]

[updated at 22:30 GMT: two typos in integrals corrected and explanation improved]

[update of July 26, 2008: \square added]

- : – : -

Remark: since

\Gamma (x)=\displaystyle\int_{0}^{+\infty} t^{x-1}e^{-t}\;dt,

by substitution, we can see that

 \displaystyle\int_{-\infty}^{\infty}e^{-x^2}\; dx=2I=\sqrt{\pi}=\Gamma \left(\dfrac{1}{2}\right) .

Find an appropriate substitution.

Plot of y=e^{-x^2}

P. S. of July 25, 2008, 12:56 GMT

Another way of evaluating  \displaystyle\int_{0}^{\infty}e^{-x^2}\; dx is shown in the following pages of Murray Spiegel, Transformadas de Laplace, Colecção Schaum, Editora McGraw Hill do Brasil, Ltda., 1971, translated from Murray Spiegel, Schaum’s Outline of Theory and Problems of Laplace Transforms, McGraw-Hill Book Co., 1965.

 

About these ads

Sobre Américo Tavares

eng. electrotécnico reformado / retired electrical engineer
Esta entrada foi publicada em Calculus, Cálculo, Demonstração, Integrais, Matemática, Math, Problem, Problemas, Proof com as etiquetas , , , . ligação permanente.

Deixar uma resposta

Preencha os seus detalhes abaixo ou clique num ícone para iniciar sessão:

Gravatar
WordPress.com Logo

Está a comentar usando a sua conta WordPress.com ( Log Out / Modificar )

Imagem do Twitter

Está a comentar usando a sua conta Twitter ( Log Out / Modificar )

Facebook photo

Está a comentar usando a sua conta Facebook ( Log Out / Modificar )

Cancelar

Connecting to %s