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Have this Integral.

$$\int_{-\infty }^\infty \frac{\cos(2x)}{(x^2+1)(x^2+4)^2}\,dx$$

Been working on similar problems but the cos bother me in this problem. Can anyone help me get started? What should I do first?

I´m given that i could use

$$\operatorname{Re}{\left ( \int_{-\infty }^\infty \frac{e^{2ix}}{(x^2+1)(x^2+4)^2} \, dx \right )} = \int_{-\infty }^\infty \frac{\cos(2x)}{(x^2+1)(x^2+4)^2}dx$$

How can the real part of this integral help me?

  • 3
    The integral under the $\operatorname{Re}$ can be computed by method of residues2011-10-06
  • 2
    No special treatment is required for the cosine. You can just substitute the pole $z_0$ into $f(z)(z-z_0)$ like you would with any other function.2011-10-06
  • 0
    The point of the hint is that the real part of $\int_a^b (u(t) + iv(t))\, dt$ is $\int_a^b u(t)\, dt$.2011-10-06

1 Answers 1