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Math Help - [SOLVED] Calculating an improper integral via complex analysis

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Calculating an improper integral via complex analysis

    I must calculate \int _0 ^{\infty} \frac{dx}{x^4+1}.
    My attempt :
    Let f(z)=\frac{1}{z^4+1}. I've calculated the poles of f to be z_i=\frac{\sqrt 2}{2}+i\frac{\sqrt2}{2}, \frac{\sqrt 2}{2}-i\frac{\sqrt2}{2}, -\frac{\sqrt 2}{2}+i\frac{\sqrt2}{2} and -\frac{\sqrt 2}{2}-i\frac{\sqrt2}{2}.

    I chose the curve \gamma _R as being the circle centered at z=0 with radius R>1.
    So I have that \int _{\gamma _R} f(z)dz=2 \pi i \sum _i Res(f,z_i).
    It remains to calculate the residues of f, which I wasn't able to do.
    I tried to express f as a Laurent series but I didn't succeed in it.
    I also tried to use the fact that the residues are worth \lim _{z \to z_i} (z-z_i)f(z) but without success in evaluating the limit.
    So how could I find the residues of f?

    Thanks in advance... and sorry! I find all this very difficult to grasp in such a few time.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by arbolis View Post
    I must calculate \int _0 ^{\infty} \frac{dx}{x^4+1}.
    My attempt :
    Let f(z)=\frac{1}{z^4+1}. I've calculated the poles of f to be z_i=\frac{\sqrt 2}{2}+i\frac{\sqrt2}{2}, \frac{\sqrt 2}{2}-i\frac{\sqrt2}{2}, -\frac{\sqrt 2}{2}+i\frac{\sqrt2}{2} and -\frac{\sqrt 2}{2}-i\frac{\sqrt2}{2}.

    I chose the curve \gamma _R as being the circle centered at z=0 with radius R>1.
    So I have that \int _{\gamma _R} f(z)dz=2 \pi i \sum _i Res(f,z_i).
    It remains to calculate the residues of f, which I wasn't able to do.
    I tried to express f as a Laurent series but I didn't succeed in it.
    I also tried to use the fact that the residues are worth \lim _{z \to z_i} (z-z_i)f(z) but without success in evaluating the limit.
    So how could I find the residues of f?

    Thanks in advance... and sorry! I find all this very difficult to grasp in such a few time.
    First note that \int_0^{\infty}\frac{\,dx}{x^4+1}=\frac{1}{2}\int_  {-\infty}^{\infty}\frac{\,dx}{x^4+1}

    Let z_1=\frac{\sqrt 2}{2}+i\frac{\sqrt2}{2}, z_2=\frac{\sqrt 2}{2}-i\frac{\sqrt2}{2}, z_3-\frac{\sqrt 2}{2}+i\frac{\sqrt2}{2} and z_4=-\frac{\sqrt 2}{2}-i\frac{\sqrt2}{2}

    If we select our closed contour to be the semicircle above the real axis, we see that we will have residues at z_1 and z_3.

    So it follows then that

    \text{Res}\!\left(f,z_1\right)=\lim_{z\to z_1}\left(z-z_1\right)\frac{1}{(z-z_1)(z-z_2)(z-z_3)(z-z_4)} =\lim_{z\to z_1}\frac{1}{(z-z_2)(z-z_3)(z-z_4)}=\frac{1}{(i\sqrt{2})(\sqrt{2})(\sqrt{2}(1+i)  )}=\frac{i+1}{-4\sqrt{2}}

    \text{Res}\!\left(f,z_3\right)=\lim_{z\to z_3}\left(z-z_3\right)\frac{1}{(z-z_1)(z-z_2)(z-z_3)(z-z_4)} =\lim_{z\to z_3}\frac{1}{(z-z_1)(z-z_2)(z-z_4)}=\frac{1}{(-\sqrt{2})(\sqrt{2}(i-1))(i\sqrt{2})}=\frac{i-1}{-4\sqrt{2}}

    Thus, by the residue theorem,

    \int_{0}^{\infty}\frac{\,dx}{x^4+1}=\frac{1}{2}\in  t_{-\infty}^{\infty}\frac{\,dx}{x^4+1}=\frac{1}{2}\lef  t(2\pi i \left(\frac{i+1}{-4\sqrt{2}}+\frac{i-1}{-4\sqrt{2}}\right)\right) =\frac{\pi}{4}\sqrt{2}

    Does this make sense?
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  3. #3
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    First note that \int_0^{\infty}\frac{\,dx}{x^4+1}=\frac{1}{2}\int_  {-\infty}^{\infty}\frac{\,dx}{x^4+1}

    Let z_1=\frac{\sqrt 2}{2}+i\frac{\sqrt2}{2}, z_2=\frac{\sqrt 2}{2}-i\frac{\sqrt2}{2}, z_3-\frac{\sqrt 2}{2}+i\frac{\sqrt2}{2} and z_4=-\frac{\sqrt 2}{2}-i\frac{\sqrt2}{2}

    If we select our closed contour to be the semicircle above the real axis, we see that we will have residues at z_1 and z_3.

    So it follows then that

    \text{Res}\!\left(f,z_1\right)=\lim_{z\to z_1}\left(z-z_1\right)\frac{1}{(z-z_1)(z-z_2)(z-z_3)(z-z_4)} =\lim_{z\to z_1}\frac{1}{(z-z_2)(z-z_3)(z-z_4)}=\frac{1}{(i\sqrt{2})(\sqrt{2})(\sqrt{2}(1+i)  )}=\frac{i+1}{-4\sqrt{2}}

    \text{Res}\!\left(f,z_3\right)=\lim_{z\to z_3}\left(z-z_3\right)\frac{1}{(z-z_1)(z-z_2)(z-z_3)(z-z_4)} =\lim_{z\to z_3}\frac{1}{(z-z_1)(z-z_2)(z-z_4)}=\frac{1}{(-\sqrt{2})(\sqrt{2}(i-1))(i\sqrt{2})}=\frac{i-1}{-4\sqrt{2}}

    Thus, by the residue theorem,

    \int_{0}^{\infty}\frac{\,dx}{x^4+1}=\frac{1}{2}\in  t_{-\infty}^{\infty}\frac{\,dx}{x^4+1}=\frac{1}{2}\lef  t(2\pi i \left(\frac{i+1}{-4\sqrt{2}}+\frac{i-1}{-4\sqrt{2}}\right)\right) =\frac{\pi}{4}\sqrt{2}

    Does this make sense?
    Yes it does, thank you so much.
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  4. #4
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    Quote Originally Posted by arbolis View Post
    I must calculate \int _0 ^{\infty} \frac{dx}{x^4+1}.
    My attempt :
    Let f(z)=\frac{1}{z^4+1}. I've calculated the poles of f to be z_i=\frac{\sqrt 2}{2}+i\frac{\sqrt2}{2}, \frac{\sqrt 2}{2}-i\frac{\sqrt2}{2}, -\frac{\sqrt 2}{2}+i\frac{\sqrt2}{2} and -\frac{\sqrt 2}{2}-i\frac{\sqrt2}{2}.

    I chose the curve \gamma _R as being the circle centered at z=0 with radius R>1.
    So I have that \int _{\gamma _R} f(z)dz=2 \pi i \sum _i Res(f,z_i).
    It remains to calculate the residues of f, which I wasn't able to do.
    I tried to express f as a Laurent series but I didn't succeed in it.
    I also tried to use the fact that the residues are worth \lim _{z \to z_i} (z-z_i)f(z) but without success in evaluating the limit.
    So how could I find the residues of f?

    Thanks in advance... and sorry! I find all this very difficult to grasp in such a few time.

    i think that it'll be easier to work if you write the roots of z^4=-1 in exponential form:

    z_i=e^{\frac{\pi i}{4}(1+2k)}\,,\,\,k=0,1,2,3 (instead exponential you can use cis, of course)

    Now, write z^4+1=\prod \limits_{i=1}^{4}(z-z_i)

    And now the evaluation of the limits for the residues is easier, imo. For instance, the residue at z_2=e^{\frac{5\pi i}{4}}=-\frac{1}{\sqrt{2}}(1+i) we get:

    \lim_{z \rightarrow z_2}(z-z_2)f(z)=\lim_{z \rightarrow z_2}\frac{1}{(z-z_0)(z-z_1)(z-z_3)}=\frac{1}{(z_2-z_0)(z_2-z_1)(z_2-z_3)} =\frac{1}{\frac{1}{2\sqrt{2}}(-2(1+i)(-2i)(-2)}=\frac{1+i}{4\sqrt{2}} (check this very carefully because there's a good chance I got it wrong)

    Tonio
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