Using the Residue Theorem for Real Integrals

liorda
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Homework Statement


I=\int_{-\infty}^{\infty} { dx \over {5x^2+6x+5}}

Homework Equations


The residue theorem.

The Attempt at a Solution


I can't use the residue theorem since the denominator has real zeros. How should I solve this?
 
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liorda said:

Homework Statement


I=\int_{-\infty}^{\infty} { dx \over {5x^2+6x+5}}

Homework Equations


The residue theorem.

The Attempt at a Solution


I can't use the residue theorem since the denominator has real zeros. How should I solve this?
Yes you can use the residue theorem, you just need to "go around" the poles, i.e. deform your usual semi-circular contour around the poles. Have a look http://www.nhn.ou.edu/~milton/p5013/chap7.pdf" and in particular at second 7.6.
 
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Maybe you should double-check that business about the real zeros.
 
awkward said:
Maybe you should double-check that business about the real zeros.
Good catch, I didn't even think to look!
 
oops. From now on, I'll never try to guess root. Quadratic equation, we meet again.

Thanks guys.
 
There is also the quaint old "complete the squares" technique:\int_{-\infty}^{\infty} \frac{dx}{5x^2+6x+5} = \frac{1}{5} \int_{-\infty}^{\infty} \frac{dx}{ (x^2+\frac{6}{5}x+\frac{9}{25}) + (1 - \frac{9}{25} ) }

= \frac{1}{5} \int_{-\infty}^{\infty} \frac{dx}{ (x+\frac{3}{5})^{2} + (\frac{4}{5})^{2} } .

I believe quadratic polynomials in the denominator never require complex-analytic techniques, though you certainly aren't forbidden to use them...
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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