# Use Complex Integrals/ Residue to evaluate $\int_0^\infty \frac{dx}{(x+1)^3 + 1}$

Use Complex Integrals/ Residue to evaluate $\int_0^\infty \frac{dx}{(x+1)^3 + 1}$

I'm not sure how to do this integration. It looks like partial fractions but I'm unsure.

• are u forced to use residue theorem? this integral is easier solved by other methods – tired Nov 26 '15 at 23:34

Generally, we can evaluate integrals of the form

$$\int_0^{\infty} dx \, f(x)$$

using a contour integral of the form

$$\oint_c dz \, f(z) \log{z}$$

As long as $f$ is sufficiently well-behaved at the origin and at infinity and in between, we have

$$-i 2 \pi \int_0^{\infty} dx \, f(x) = i 2 \pi \sum_k \operatorname*{Res}_{z=z_k} f(z) \log{z_k}$$

In this case

$$f(z) = \frac1{(1+z)^3+1}$$

so that, for $k \in \{0,1,2\}$:

$$z_k = -1 + e^{i (2 k+1) \pi/3} = -\left [1-\cos{(2 k+1) \frac{\pi}{3}}\right ] + i \sin{(2 k+1) \frac{\pi}{3}}$$

Thus,

$$\int_0^{\infty} \frac{dx}{(1+x)^3+1} = -\sum_{k=0}^2 \frac{\log{z_k}}{3 e^{i 2 (2 k+1) \pi/3}}$$

where

\begin{align}\log{z_k} &= \log{|z_k|} + i \arg{z_k} \\ &= \log{2 \left | \sin{(2 k+1) \frac{\pi}{6}} \right |} - i \arctan{\cot{(2 k+1) \frac{\pi}{6}}} \\ &= \log{2} - i \frac{\pi}{2} + \log{\sin{(2 k+1) \frac{\pi}{6}}} + i (2 k+1) \frac{\pi}{6}\end{align}

so that

\begin{align}\sum_{k=0}^2 \frac{\log{z_k}}{3 e^{i 2 (2 k+1) \pi/3}} &= -i \frac13 \frac{\pi}{3} e^{-i 2 \pi/3} + \frac13 \log{2} e^{-i 2 \pi} + i \frac13 \frac{\pi}{3} e^{i 2 \pi/3}\\ &= -\frac{2 \pi}{9} \sin{\frac{2 \pi}{3}} + \frac13 \log{2}\end{align}

Finally, we may conclude that

$$\int_0^{\infty} \frac{dx}{(1+x)^3+1} = \frac{\pi}{3 \sqrt{3}} - \frac13 \log{2}$$

This agrees with a numerical integral in Mathematica v 9.0.

We have $$I = \int_0^{\infty} \dfrac{dx}{(1+x)^3+1} = \int_1^{\infty} \dfrac{dx}{x^3+1} = \int_1^0 \dfrac{-dx/x^2}{1/x^3+1} = \int_0^1 \dfrac{xdx}{1+x^3}$$ Hence, $$I = \int_0^1\dfrac{x+1}{3(x^2-x+1)}dx - \int_0^1\dfrac{dx}{3(x+1)} = \dfrac{\pi}{3\sqrt{3}}-\dfrac{\log2}3$$

$$I=\int_{0}^{+\infty}\frac{dx}{(x+1)^3+1}=\int_{1}^{+\infty}\frac{dx}{x^3+1}=\int_{0}^{1}\frac{x}{1+x^3}\,dx$$ The roots of $1+x^3$ lie at $-1,\xi=\frac{1+i\sqrt{3}}{2},\bar{\xi}=\frac{1-i\sqrt{3}}{2}$ and they are simple. Since: $$\text{Res}\left(\frac{x}{1+x^3},x=-1\right) = -\frac{1}{3}$$ the partial fraction decomposition of $\frac{x}{1+x^3}$ is given by: $$\frac{x}{1+x^3} = -\frac{1}{3(x+1)}+\frac{1+x}{3(1-x+x^2)}$$ and: $$I = \frac{1}{2}-\frac{1}{5}+\frac{1}{8}-\frac{1}{11}+\ldots = \color{red}{-\frac{\log(2)}{3}+\frac{\pi}{3\sqrt{3}}}.$$

• Jack, no offense, but your previous answer was wrong and your correction was merely a change to the correct answer provided by others. I think you really should show correction to a step as well. For example, are you sure your partial fractions decomposition is correct? Hint: @Leg got it right. – Ron Gordon Nov 27 '15 at 10:44
• @RonGordon: no offense took, but I simply forgot to divide by three, then fixed it wrong. The decomposition $$\frac{x}{1+x^3}=-\frac{1}{3(x+1)}+\frac{1+x}{3(1-x+x^2)}$$ is now correct. – Jack D'Aurizio Nov 27 '15 at 10:48
• Jack, with you I figured it was something as trivial as that. – Ron Gordon Nov 27 '15 at 10:56

HINT:

$$\int\frac{1}{\left(x+1\right)^3+1}\space\text{d}x=$$

Substitute $u=x+1$ and $\text{d}u=\text{d}x$:

$$\int\frac{1}{u^3+1}\space\text{d}u=$$ $$\int\left(\frac{2-u}{3(u^2-u+1)}+\frac{1}{3(u+1)}\right)\space\text{d}u=$$ $$\frac{1}{3}\int\frac{2-u}{u^2-u+1}\space\text{d}u+\frac{1}{3}\int\frac{1}{u+1}\space\text{d}u=$$ $$\frac{1}{3}\int\left(\frac{3}{2(u^2-u+1)}-\frac{2u-1}{2(u^2-u+1)}\right)\space\text{d}u+\frac{1}{3}\int\frac{1}{u+1}\space\text{d}u=$$ $$-\frac{1}{6}\int\frac{2u-1}{u^2-u+1}\space\text{d}u+\frac{1}{2}\int\frac{1}{u^2-u+1}\space\text{d}u+\frac{1}{3}\int\frac{1}{u+1}\space\text{d}u=$$

Substitute $s=u^2-u+1$ and $\text{d}s=(2u-1)\space\text{d}u$:

$$-\frac{1}{6}\int\frac{1}{s}\space\text{d}s+\frac{1}{2}\int\frac{1}{u^2-u+1}\space\text{d}u+\frac{1}{3}\int\frac{1}{u+1}\space\text{d}u=$$ $$-\frac{\ln|s|}{6}+\frac{1}{2}\int\frac{1}{u^2-u+1}\space\text{d}u+\frac{1}{3}\int\frac{1}{u+1}\space\text{d}u=$$ $$-\frac{\ln|s|}{6}+\frac{1}{2}\int\frac{1}{\left(u-\frac{1}{2}\right)^2+\frac{3}{4}}\space\text{d}u+\frac{1}{3}\int\frac{1}{u+1}\space\text{d}u=$$

Substitute $p=u-\frac{1}{2}$ and $\text{d}p=\text{d}u$:

$$-\frac{\ln|s|}{6}+\frac{1}{2}\int\frac{1}{p^2+\frac{3}{4}}\space\text{d}u+\frac{1}{3}\int\frac{1}{u+1}\space\text{d}u=$$ $$-\frac{\ln|s|}{6}+\frac{1}{2}\int\frac{4}{\frac{4p^2}{3}+1}\space\text{d}u+\frac{1}{3}\int\frac{1}{u+1}\space\text{d}u=$$ $$-\frac{\ln|s|}{6}+\frac{2}{3}\int\frac{1}{\frac{4p^2}{3}+1}\space\text{d}u+\frac{1}{3}\int\frac{1}{u+1}\space\text{d}u=$$

Substitute $2=\frac{2p}{\sqrt{3}}$ and $\text{d}w=\frac{2}{\sqrt{3}}\space\text{d}p$:

$$-\frac{\ln|s|}{6}+\frac{1}{\sqrt{3}}\int\frac{1}{w^2+1}\space\text{d}u+\frac{1}{3}\int\frac{1}{u+1}\space\text{d}u=$$ $$-\frac{\ln\left|s\right|}{6}+\frac{\arctan\left(w\right)}{\sqrt{3}}+\frac{1}{3}\int\frac{1}{u+1}\space\text{d}u=$$

Substitute $v=u+1$ and $\text{d}v=\text{d}u$:

$$-\frac{\ln\left|s\right|}{6}+\frac{\arctan\left(w\right)}{\sqrt{3}}+\frac{1}{3}\int\frac{1}{v}\space\text{d}v=$$ $$-\frac{\ln\left|s\right|}{6}+\frac{\arctan\left(w\right)}{\sqrt{3}}+\frac{\ln\left|v\right|}{3}+\text{C}=$$ $$-\frac{\ln\left|s\right|}{6}+\frac{\arctan\left(\frac{2p}{\sqrt{3}}\right)}{\sqrt{3}}+\frac{\ln\left|v\right|}{3}+\text{C}=$$ $$-\frac{\ln\left|s\right|}{6}+\frac{\arctan\left(\frac{2\left(u-\frac{1}{2}\right)}{\sqrt{3}}\right)}{\sqrt{3}}+\frac{\ln\left|v\right|}{3}+\text{C}=$$ $$-\frac{\ln\left|u^2-u+1\right|}{6}+\frac{\arctan\left(\frac{2\left(u-\frac{1}{2}\right)}{\sqrt{3}}\right)}{\sqrt{3}}+\frac{\ln\left|u+1\right|}{3}+\text{C}=$$ $$-\frac{\ln\left|(x+1)^2-(x+1)+1\right|}{6}+\frac{\arctan\left(\frac{2\left((x+1)-\frac{1}{2}\right)}{\sqrt{3}}\right)}{\sqrt{3}}+\frac{\ln\left|(x+1)+1\right|}{3}+\text{C}=$$

$$\frac{1}{6}\left(-\ln\left|x^2+x+1\right|+2\ln\left|x+2\right|+2\sqrt{3}\arctan\left(\frac{2x+1}{\sqrt{3}}\right)\right)+\text{C}$$

$$\lim_{n\to\infty}\left[\frac{1}{6}\left(-\ln\left|x^2+x+1\right|+2\ln\left|x+2\right|+2\sqrt{3}\arctan\left(\frac{2x+1}{\sqrt{3}}\right)\right)\right]_{0}^{n}=$$ $$\lim_{n\to\infty}\left(\frac{1}{18}\left(-3\ln|n^2+n+1|+6\ln|n+2|+6\sqrt{3}\arctan\left(\frac{2n+1}{\sqrt{3}}\right)-\pi\sqrt{3}-\ln(64)\right)\right)$$