Integral u^2/(1+u^4)

  • Thread starter Aquinox
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  • #1
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Homework Statement


The initial problem is to calculate
[tex]\int_{-\infty}^{\infty}\cos(x^{2})dx[/tex] using
[tex]t=x^{2}[/tex]
and then
[tex]t^{-\frac{1}{2}}=\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-tu^{2}}du[/tex]

Homework Equations





The Attempt at a Solution


I have, by transformation and use of the symmetry of both integrals come to
[tex]\int_{0}^{\infty}\frac{u^{2}}{1+u^{4}}du[/tex]
which is easily solveable using Mathematica.
Alas the solution provided by mathematica is absolutely non-obvious to me.

Any good idea for a substitution and/or other ways?
I've tried splitting the fraction into smaller parts, but got an integral with i in it, which is not allowed as this is real calculus.

Thanks in advance
 

Answers and Replies

  • #2
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[tex] \frac{u^2}{1+u^4}=\frac{1}{4\sqrt{2}}\left(\frac{2 u-\sqrt{2}}{u^2-\sqrt{2} u+1}-\frac{2 u+\sqrt{2}}{u^2+\sqrt{2} u+1}+\frac{2 \sqrt{2}}{\left(1-\sqrt{2} u\right)^2+1}+\frac{2 \sqrt{2}}{\left(\sqrt{2} u+1\right)^2+1}\right)\right)[/tex]

The first two terms integrate to logs the next to to arctans, then take limits.
 
  • #3
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That is the answer I, too, have derived from the solutions Mathematica gave me.

What I need to know is which method to use to get from the left side to the right side.
That is a step non-obvious to me and it has to be.
 
  • #4
459
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Well it would be helpful if you had written the problem a bit clearer. In my post the right side is just partial fractions of the thing on the left side. If you need to work out the first integral, once you show its the same as the Gaussian you can find the value of that integral in a much easier way than using those partial fractions.
 
  • #5
10
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Sorry.

The way hinted to me in the paper is that we have to take that t^(-1/2) definition and use fubini's thm.

This leads to the aforementioned


[tex]
\int_{0}^{\infty}\frac{u^{2}}{1+u^{4}}du
[/tex]

I am perfectly able to integrate the RHS of the equation you've posted, but here is my point:

In this equation:

[tex]
\frac{u^2}{1+u^4}=\frac{1}{4\sqrt{2}}\left(\frac{2 u-\sqrt{2}}{u^2-\sqrt{2} u+1}-\frac{2 u+\sqrt{2}}{u^2+\sqrt{2} u+1}+\frac{2 \sqrt{2}}{\left(1-\sqrt{2} u\right)^2+1}+\frac{2 \sqrt{2}}{\left(\sqrt{2} u+1\right)^2+1}\right)\right)
[/tex]

which method does one use to produce the RHS from the LHS (u^2/(1+u^4))? (sic!)

As long as I can not state that step the solution will not be valid.(as not all steps will be listed)
 
  • #6
Char. Limit
Gold Member
1,204
14
It's just partial fraction decomposition on u^4+1.
 

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