I am attempting to solve the following integral via contour methods. It is from a past exam paper to which I have solutions, but I do not understand a key part of the method.
$$ I = \int_{1}^{\infty} \frac{x(x^2+3)}{(x^2+1)(x^2+4)\sqrt{x^2-1}} dx $$
The contour I am using for the integral looks like this, with the green squiggle denoting the branch cut:
I am comfortable with establishing that the contributions from the circular parts of the contour vanish, but I am convinced that the contributions from the straight sections (1) and (3) cancel with each other, and likewise for (5) and (7). Obviously this cannot be the case since the residues do not cancel, but given that the integrand is odd with respect to the real axis and we are integrating in the same direction for sections (1) and (3), why do they not directly cancel?
If someone can see the lapse in my understanding that would be great, thanks!
Knowing whether or not contour integrals over line segments going in the opposite (or same) direction depends on what you define your arguments to be.
Let $\displaystyle I = \int_{1}^{\infty}\frac{x\left(x^{2}+3\right)}{\left(x^{2}+1\right)\left(x^{2}+4\right)\sqrt{x^{2}-1}}dx$, let $\displaystyle f(z) = \frac{z\left(z^{2}+3\right)}{\left(z^{2}+1\right)\left(z^{2}+4\right)\sqrt{z^{2}-1}}$, and let $\displaystyle g(z) = \frac{z\left(z^{2}+3\right)}{\left(z^{2}+1\right)\left(z^{2}+4\right)}$. By using the complex definition of logarithms, we have
$$f(z) = \frac{g(z)}{\exp\left(\frac{1}{2}\ln\left|z+1\right|+\frac{i}{2}\operatorname{arg}\left(z+1\right)\right)\exp\left(\frac{1}{2}\ln\left|z-1\right|+\frac{i}{2}\operatorname{arg}\left(z-1\right)\right)}.$$
Just like in the picture above, let $\operatorname{arg}(z+1) \in (-\pi, \pi)$ and $\operatorname{arg}(z-1) \in (0, 2\pi)$. Let $r$ be the length of the small gap between a line segment and whichever branch cut is closest to it.
Parameterizing the integral over $C_1$ (the contour labeled "$1$" with the circle around it), we get
$$ \int_{C_1}f(z)dz = \int_{1+r+ir}^{R+ir}f(z)dz = \int_{1+r}^{R}f(x+ir)d(x+ir)\\ $$
which is
$$ \int_{1+r}^{R}\frac{g(x+ir)dx}{\exp\left(\frac{1}{2}\ln\left|x+ir+1\right|+\frac{i}{2}\operatorname{arg}\left(x+ir+1\right)\right)\exp\left(\frac{1}{2}\ln\left|x+ir-1\right|+\frac{i}{2}\operatorname{arg}\left(x+ir-1\right)\right)}. $$
Taking $r \to 0^+$ and $R \to \infty$, we get the above expression to converge to
$$ \int_{1}^{\infty}\frac{g(x)}{\exp\left(\frac{1}{2}\ln\left|x+1\right|+\frac{i}{2}\cdot0\right)\exp\left(\frac{1}{2}\ln\left|x-1\right|+\frac{i}{2}\cdot0\right)}dx = I. $$
Parameterizing the integral over $C_3$, we get
$$ \int_{C_3}f(z)dz = \int_{-R+ir}^{-1-r+ir}f(z)dz = \int_{R}^{1+r}f(-x+ir)d(-x+ir)\\ $$
which is
$$ -\int_{R}^{1+r}\frac{g(-x+ir)dx}{\exp\left(\frac{1}{2}\ln\left|-x+ir+1\right|+\frac{i}{2}\operatorname{arg}\left(-x+ir+1\right)\right)\exp\left(\frac{1}{2}\ln\left|-x+ir-1\right|+\frac{i}{2}\operatorname{arg}\left(-x+ir-1\right)\right)}. $$
Taking $r \to 0^+$ and $R \to \infty$, we get the above expression to converge to
$$ \int_{\infty}^{1}\frac{g(x)}{\exp\left(\frac{1}{2}\ln\left|-x+1\right|+\frac{i}{2}\cdot \pi\right)\exp\left(\frac{1}{2}\ln\left|-x-1\right|+\frac{i}{2}\cdot \pi\right)}dx = I. $$
Parameterizing the integral over $C_5$, we get
$$ \int_{C_5}f(z)dz = \int_{-1-r-ir}^{-R-ir}f(z)dz = \int_{1+r}^{R}f(-x-ir)d(-x-ir)\\ $$
which is
$$ -\int_{1+r}^{R}\frac{g(-x-ir)dx}{\exp\left(\frac{1}{2}\ln\left|-x-ir+1\right|+\frac{i}{2}\operatorname{arg}\left(-x-ir+1\right)\right)\exp\left(\frac{1}{2}\ln\left|-x-ir-1\right|+\frac{i}{2}\operatorname{arg}\left(-x-ir-1\right)\right)}. $$
Taking $r \to 0^+$ and $R \to \infty$, we get the above expression to converge to
$$ \int_{1}^{\infty}\frac{g(x)}{\exp\left(\frac{1}{2}\ln\left|-x+1\right|+\frac{i}{2}\cdot (-\pi)\right)\exp\left(\frac{1}{2}\ln\left|-x-1\right|+\frac{i}{2}\cdot \pi\right)}dx = I. $$
Parameterizing the integral over $C_7$, we get
$$ \int_{C_7}f(z)dz = \int_{R-ir}^{1+r-ir}f(z)dz = \int_{R}^{1+r}f(x-ir)d(x-ir)\\ $$
which is
$$ \int_{R}^{1+r}\frac{g(x-ir)dx}{\exp\left(\frac{1}{2}\ln\left|x-ir+1\right|+\frac{i}{2}\operatorname{arg}\left(x-ir+1\right)\right)\exp\left(\frac{1}{2}\ln\left|x-ir-1\right|+\frac{i}{2}\operatorname{arg}\left(x-ir-1\right)\right)}. $$
Taking $r \to 0^+$ and $R \to \infty$, we get the above expression to converge to
$$ \int_{\infty}^{1}\frac{g(x)}{\exp\left(\frac{1}{2}\ln\left|x+1\right|+\frac{i}{2}\cdot0\right)\exp\left(\frac{1}{2}\ln\left|x-1\right|+\frac{i}{2}\cdot 2\pi\right)}dx = I. $$