Whenever I tried to do, it failed. Is there anyone to help? $$\int_0^\infty \frac{dx}{1+x^6}$$
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1What methods can you use? The integral is straightforward in principle but very messy if you use real partial fractions. It is probably no harder to use complex methods, if you have studied these. – David Nov 09 '14 at 23:43
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Here is a relevant MSE link. – Marko Riedel Nov 09 '14 at 23:49
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@user153012 it is only a duplicate if topy's aim is to use real analysis techniques as the title says. – dustin Nov 10 '14 at 00:28
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Hint: find the roots of $x^6=−1$ in the complex plane, then by the factor theorem $$x^6+1=(x-x_1)(x-x_2)\cdots(x-x_6)=(x^2+b_1x+c_1)(x^2+b_2x+c_2)(x^2+b_3x+c_3)$$ And apply partial fractions.
AsdrubalBeltran
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Since your integral is even, we can write it as $$ \frac{1}{2}\int_{-\infty}^{\infty}\frac{dz}{z^6 + 1} $$ Let $z = re^{i\theta}$. Then $z^6 = r^6e^{6i\theta} = e^{i(\pi + 2\pi k)}$. So $r = 1$ and $\theta = \frac{\pi}{6} + \frac{\pi k}{3}$. Also, denote $g(z) = z^6 + 1$ so $g'(z) = 6z^5$ which is only zero iff $z = 0$; therefore, $1/g$ only has simple poles.
- Determine poles in upper half plane.
- $\frac{1}{2}\int = \pi i\sum_{z_j\in\text{UHP}}\text{Res}_{z_j}\frac{1}{g'(z)}$
This should all be fairly straight forward.
dustin
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2Simpler method: integrate $f(z)=1/(z^6+1)$ around the boundary of the circular sector $0<r<R$, $0<\theta<\frac\pi3$. Then you only have to deal with one pole. It also generalises easily to $1/(z^n+1)$. – David Nov 10 '14 at 00:10
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It's not too bad in this case. But for a general $n$ it gets a bit messy. – David Nov 10 '14 at 00:29
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How to calculate $\displaystyle\int_0^\infty\frac{dx}{1+x^6}$ ?
By letting $t=\dfrac1{1+x^6}$ and recognizing the expression of the beta function in the new
integral, then using Euler's reflection formula for the $\Gamma$ function. All integrals of the
form $\displaystyle\int_0^\infty\frac{x^k}{a^n+x^n}~dx~$ can be evaluated in this manner.
Lucian
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