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

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

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

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

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

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

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

    So it follows then that

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

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

    Thus, by the residue theorem,

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

    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 $\displaystyle \int_0^{\infty}\frac{\,dx}{x^4+1}=\frac{1}{2}\int_ {-\infty}^{\infty}\frac{\,dx}{x^4+1}$

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

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

    So it follows then that

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

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

    Thus, by the residue theorem,

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

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

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

    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 $\displaystyle z^4=-1$ in exponential form:

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

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

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

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

    Tonio
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