- #1
Nikitin
- 735
- 27
[solved]Easy complex analysis question
Hi. In the complex plain, since y = 0 (in z=x+iy) at the x axis, shouldn't the following be true? :
##y=0##
[tex] \int_{-\infty}^{\infty} \frac{\cos(ax)}{x^2+2x+5} dx = \int_{-\infty}^{\infty} \frac{e^{iaz}}{z^2+2z+5} dz = \int_{-\infty}^{\infty} f(z) dz[/tex]
Or does it only go like this:
##y=0##
[tex] \int_{-\infty}^{\infty} \frac{\cos(ax)}{x^2+2x+5} dx = Re(\int_{-\infty}^{\infty} \frac{e^{iaz}}{z^2+2z+5} dz)= Re( \int_{-\infty}^{\infty} f(z) dz)[/tex]
I am asking because I need to find the value of the integral ##\int_{-\infty}^{\infty}\frac{\cos(ax)}{x^2+2x+5} dx## using the residue theorem, but I don't know whether I need to use the residue of ##Re(\int_{-\infty}^{\infty} f(z) dz)## or ##\int_{-\infty}^{\infty} f(z) dz##. The problem is that if I use the latter, my logic seems coherent but I get a residue which has both a real and imaginary part, which hints that I should find the residue of the former instead.
Hi. In the complex plain, since y = 0 (in z=x+iy) at the x axis, shouldn't the following be true? :
##y=0##
[tex] \int_{-\infty}^{\infty} \frac{\cos(ax)}{x^2+2x+5} dx = \int_{-\infty}^{\infty} \frac{e^{iaz}}{z^2+2z+5} dz = \int_{-\infty}^{\infty} f(z) dz[/tex]
Or does it only go like this:
##y=0##
[tex] \int_{-\infty}^{\infty} \frac{\cos(ax)}{x^2+2x+5} dx = Re(\int_{-\infty}^{\infty} \frac{e^{iaz}}{z^2+2z+5} dz)= Re( \int_{-\infty}^{\infty} f(z) dz)[/tex]
I am asking because I need to find the value of the integral ##\int_{-\infty}^{\infty}\frac{\cos(ax)}{x^2+2x+5} dx## using the residue theorem, but I don't know whether I need to use the residue of ##Re(\int_{-\infty}^{\infty} f(z) dz)## or ##\int_{-\infty}^{\infty} f(z) dz##. The problem is that if I use the latter, my logic seems coherent but I get a residue which has both a real and imaginary part, which hints that I should find the residue of the former instead.
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