Solve integral using residue theorem

[Siberion]

Homework Statement

Considering the following integral,

$$I = \int^\infty_{-\infty} \frac{x^2}{1+x^4}$$

I can rewrite it as a complex contour integral as:

$$\oint^{}_{C} \frac{z^2}{1+z^4}$$

where the contour C is a semicircle on the half-upper plane with a radius which extends to infinity. I can use the residue theorem and get the result by noticing the singularities of the function, which in this case are just simple poles.

However, I'm not sure about what is the correct approach when the integral goes from 0 to infinity, i.e.

$$I = \int^\infty_{0} \frac{x^2}{1+x^4}$$

I was tempted to use the same contour of integration, but that would give me the same result ,so I really wish to know what is the correct contour to use in this kind of case.

Given that the integrand is a pair function, would the result be just 1/2 of the integral from -∞ to ∞?

Any help would be greatly appreciated.

Thanks for your time.

 

[Dick]

Homework Statement

Considering the following integral,

$$I = \int^\infty_{-\infty} \frac{x^2}{1+x^4}$$

I can rewrite it as a complex contour integral as:

$$\oint^{}_{C} \frac{z^2}{1+z^4}$$

where the contour C is a semicircle on the half-upper plane with a radius which extends to infinity. I can use the residue theorem and get the result by noticing the singularities of the function, which in this case are just simple poles.

However, I'm not sure about what is the correct approach when the integral goes from 0 to infinity, i.e.

$$I = \int^\infty_{0} \frac{x^2}{1+x^4}$$

I was tempted to use the same contour of integration, but that would give me the same result ,so I really wish to know what is the correct contour to use in this kind of case.

Given that the integrand is a pair function, would the result be just 1/2 of the integral from -∞ to ∞?

Any help would be greatly appreciated.

Thanks for your time.

If a "pair" function is an "even" function, i.e. f(-x)=f(x), then yes, the integral from 0 to infinity is (1/2) of the integral from -infinity to infinity.

 

[Siberion]

If a "pair" function is an "even" function, i.e. f(-x)=f(x), then yes, the integral from 0 to infinity is (1/2) of the integral from -infinity to infinity.

Haha, my apologies, I got confused with the word commonly used in spanish "par". In english, it would indeed be an even function.