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I am trying to find $\int_0^\infty \dfrac{dx}{1+x^4}$ by setting it equal to $\dfrac{1}{2}\oint_C \dfrac{dz}{1+z^4}$ and solving that. By a computer program I've calculated it to be $\approx 1.11072$; I am not getting that answer though. I factor it into $\dfrac{1}{2}\oint_C \dfrac{dz}{(z-1)(z+1)(z^2 + 1)}$; there is a simple pole at $z = 1$ and a 4th order pole at $z = i$; I find the residue at 1 to be $\dfrac{1}{4}$ and at $i$ to be $\dfrac{-1}{4}$; the second pole is found by L'Hopital's rule. Could someone tell me what I'm missing? Thanks

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$(z - 1)(z + 1)(z^2+1) = z^4 - 1$, not $z^4 + 1$. –  achille hui Feb 12 '13 at 2:15
    
Aside: the rational function $$\frac{1}{(z-1)(z+1)(z^2+1)}$$ only has a simple pole at $z=i$. Also note $z^2+1 = (z+i)(z-i)$. –  Hurkyl Feb 12 '13 at 4:13
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1 Answer

up vote 1 down vote accepted

$$z^4+1=(z^2+i)(z^2-i)=(z-w_1)(z-w_2)(z-u_1)(z-u_2)$$

where

$$w_k=\pm\frac{1}{\sqrt 2}(1-i)\;\;\;,\;\;\;u_k=\pm\frac{1}{\sqrt 2}(1+i)\,\,\,,\,\,k=1,2$$

Thus the function $\,\displaystyle{f(z):=\frac{1}{z^4+1}}\,$ has two simple poles in the upper half plane:

$$Res_{z=\frac{1}{\sqrt 2}(-1+i)}(f)=\lim_{z\to-\frac{1}{\sqrt 2}(-1+i)}\,\left(z+\frac{1}{\sqrt 2}(-1+i)\right)f(z)\stackrel{L'Hospital}=$$

$$=\frac{1}{4\left[\frac{1}{\sqrt 2}\left(-1+i\right)\right]^3}=\frac{1}{2\sqrt 2(1+i)}$$

$$Res_{z=\frac{1}{\sqrt 2}(1+i)}(f)=\lim_{z\to\frac{1}{\sqrt 2}(1+i)}\,\left(z-\frac{1}{\sqrt 2}(1+i)\right)f(z)\stackrel{L'Hospital}=$$

$$=\frac{1}{4\left[\frac{1}{\sqrt 2}\left(1+i\right)\right]^3}=\frac{1}{2\sqrt 2(-1+i)}$$

Choosing now the contour

$$C_R:=[-R,R]\cup\gamma_R:=\{z\in\Bbb C\;;\;z=Re^{it}\,,\,\,0\le t\le \pi\}\;\;,\;\;0<<R\in\Bbb R$$

we thus get by Cauchy's Residues Theorem:

$$\oint_{C_R}f(z)\,dz=2\pi i\left(-\frac{i}{2\sqrt 2}\right)=\frac{\pi}{\sqrt 2}$$

Now two tasks for you:

1) Prove that the above integral approaches zero when $\,R\to\infty\,$ on $\,\gamma_R\,$

2) On $\,[-R,R]\,$ , the integral approaches twice your integral (use the fact that the integrand function is even)

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