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Improper integral using residues

  • #1
114
0

Homework Statement


[tex]\int\limits^{ +\infty }_{0}\frac{ \sqrt{x} \mbox{d} x }{ x^2+1 }[/tex]

The Attempt at a Solution


I make substitution [tex]\sqrt{x}=t[/tex] and then
[tex]\int^{ +\infty }_{0}\frac{ t^2 \mbox{d} t }{ t^4+1 }[/tex]
and now this function is odd, so I make a half circle and count residues yeah?
 

Answers and Replies

  • #2
211
0
Since you have [tex]\sqrt{x}[/tex] I believe you have a branch cut along the x-axes. Your integral should be:

[tex]\oint\frac{z^{1/2}}{z^2+1}dz[/tex]

I'd try to solve it and I'll get back to you :D

EDIT: Yep I'm right, solved it :D

You have a branch cut along the x-axes.

The result is [tex]\frac{\pi}{\sqrt{2}}[/tex]

If you want I can scan you the procedure ^^

Plus how did you get to the conclusion that the function under the integral is odd? :\
 
Last edited:
  • #3
211
0
Also the integral [tex]\int^{ +\infty }_{0}\frac{ t^2 \mbox{d} t }{ t^4+1 }=\frac{\pi }{2 \sqrt{2}}[/tex]

While [tex]\int^{ +\infty }_{0}\frac{ t^2 \mbox{d} t }{ t^4+1 }=\frac{\pi }{\sqrt{2}}[/tex]

That's why you need the branch cut...
 

Attachments

  • #4
uart
Science Advisor
2,776
9
Also the integral [tex]\int^{ +\infty }_{0}\frac{ t^2 \mbox{d} t }{ t^4+1 }=\frac{\pi }{2 \sqrt{2}}[/tex]

While [tex]\int^{ +\infty }_{0}\frac{ t^2 \mbox{d} t }{ t^4+1 }=\frac{\pi }{\sqrt{2}}[/tex]

That's why you need the branch cut...
That was just a small mistake by the OP during the substitution where he left out a 2 ([itex] dx = 2 t \, dt[/itex]). The correct integral (after substitution) should have been.

[tex]\int^{ +\infty }_{0}\frac{ 2 t^2 \mbox{d} t }{ t^4+1 }=\frac{\pi }{\sqrt{2}}[/tex]

So actually no branch cut was required.

BTW. Normally in the homework section it's better to give hints and advice on how to proceed rather than answers or worked solutions.
 
  • #5
211
0
Yeah, I had some time to spend...
 

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