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Real integrals using complex analysis

  1. Apr 10, 2014 #1
    1. The problem statement, all variables and given/known data
    After successfully solving a lot of integrals I gathered 4 ugly ones that I can not solve:

    a) ## \int _{-\infty} ^\infty \frac{cos(2x)}{x^4+1}dx##

    b) ##\int _0 ^\infty \frac{dx}{1+x^3}##

    c) ##\int _0 ^\infty \frac{x^2+1}{x^4+1}dx##

    d) ##\int _0 ^{2\pi } \frac{d\varphi }{a+cos(\varphi )}##


    2. Relevant equations



    3. The attempt at a solution

    I simply don't know what to do with integrals ##\int _0 ^\infty##. Have no idea! Please help!

    I can only show my work and a) and d):

    a)

    ##\int_{-\infty }^{\infty }\frac{cos(2z)}{z^4+1}dz=\int_{-R }^{R }\frac{cos(2x)}{x^4+1}dx+\int _\gamma\frac{cos(2z)}{z^4+1}dz## where ##R->\infty##.

    Last integral is for ##z=Re^{i\varphi}## where ##R->\infty## clearely ##0##, therefore only one integral still remains:

    ##\int_{-\infty }^{\infty }\frac{cos(2z)}{z^4+1}dz##

    ##\int_{-\infty }^{\infty }\frac{cos(2z)}{z^4+1}dz=2\pi i \sum Res(f,a)##

    ##z^4+1=0## gives me ##z_1=e^{i\pi /4}##, ##z_2=e^{i3pi /4}##, ##z_3=e^{i5\pi /4}## and ##z_4=e^{i7\pi /4}## singularity points which are all poles of order 1.

    Therefore for all singularity points ##Res_{z=z_i} \frac{f(z)}{g(z)}=\frac{f(z_i)}{g^{'}(z_i)}=\frac{cos(2z_i)}{4z_i^3}##.

    Now this is where it all stops for me. How much is for example ##\frac{cos(2e^{i5\pi /4})}{3e^{i15\pi /4}}## ????

    d)
    ##\int _0 ^{2\pi } \frac{d\varphi }{a+cos(\varphi )}##

    Hmmm... Can I say that ##cos(\varphi )=\frac{z+\bar{z}}{2}##?
     
    Last edited: Apr 10, 2014
  2. jcsd
  3. Apr 10, 2014 #2

    HallsofIvy

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    Staff Emeritus
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    Integrals of functions of complex variables are typically path integrals around closed paths so that the integral is the sum of the residues at all poles inside the paths. In none of these have you said what closed paths you are using! For example, because (a) has poles at four points, two in the upper half plane and two in the lower, I might be inclined to integrate along the real axis, from -R to R, then take the integral around the semi-circle in the upper half plane from r= R to r= -R (I presume that is your "[itex]\gamma[/itex]" but you should say so). As long as R is larger than 1, you will have both pole inside the path. And if you can prove that the integral around the semi-circle goes to 0 as R goes to infinity, that will be equal to the desired integral.

    I don't understand why you switched from "2z" to "cos(2z)" in (a).
     
  4. Apr 10, 2014 #3
    Because it was a typo. The problem is with ##cos(2z)##. I have edited the first post now. And yes, you are correct, that is my ##\gamma ##.
     
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