As an application of Cauchy's Residue Theorem, integrals of the form $\displaystyle \int_{-\infty}^{\infty} f(x) ~d x$, when the bounds are $0$ to $\infty$ and the integrand has finite number of isolated singularities, can we always use a large enough quarter circle contour in the 1st quadrant to find the value of the given integral (rather than using a large enough half circle) with the property
$$\int_{0}^{\infty} f(x) ~d x=\frac{1}{2} \int_{-\infty}^{\infty} f(x) ~d x ,$$
when the integrand is an even function ?
For example, using both methods, I can establish $\displaystyle \int_{0}^{\infty} \dfrac{x^{2}+1}{x^{4}+1} d x=\dfrac{\pi}{\sqrt{2}}$ but the quarter circle contour didn't give the required answer in $\displaystyle \int_{0}^{\infty} \dfrac{x^{6}}{\left(x^{4}+1\right)^{2}} d x=\dfrac{3 \pi \sqrt{2}}{16}$. Wonder what's the reason for the discrepancy.
Best Answer
Of course you can use the quarter circle contour to compute integral. But I generally don't recommend this because it's not always the easiest solution. You can flexibly choose contours to solve different integral.
Example
As the OP's example, compute $\displaystyle{\int_0^{\infty}\frac{x^6}{(x^4+1)^2}}\, \mathrm{d}x$ via the following contour.
Let $f(z) = \dfrac{z^6}{(z^4+1)^2}$, it is to show that
$$ \int_{C_{R}} f(z) \,\mathrm{d}z = \int_{C_{R}}\frac{z^6}{(z^4+1)^2}\,\mathrm{d}z=0. $$
This is beacuse $\displaystyle{\lim_{R\to\infty}R\mathrm{e}^{i\theta}f(R\mathrm{e}^{i\theta}) =0}$. There is only one isolated singularity $z=\mathrm{e}^{\pi i/4}$ in the contour. With Residue theorem, we can find
$$ \int_0^{\infty}f(z)\,\mathrm{d}z + \int_{C_{R}}f(z) \, \mathrm{d}z + \int_{i\infty}^{0} f(z) \, \mathrm{d}z = 2\pi i \cdot \mathrm{Res} f(\mathrm{e}^{\pi i/4}), $$
So $$ (1+i) \int_0^{\infty}\frac{z^6}{(z^4+1)^2}\, \mathrm{d}z = 2\pi i \cdot\mathrm{Res} f(\mathrm{e}^{\pi i/4}) = 2\pi i \cdot \frac{-3}{16}\left(-\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i\right), $$
thus the final result is $\displaystyle{\int_0^{\infty}\frac{x^6}{(x^4+1)^2}\, \mathrm{d}x = \frac{3\pi\sqrt{2}}{16}}$.
Using the quarter circle contour is not always the easiest solution. For this type of integral problem (the bounds are $0$ to $\infty$). You can choose the semicirlce contour when the integrand is even.
$$ \int_0^{\infty} f(z)\, \mathrm{d}z = \frac{1}{2} \int_{-\infty}^{\infty}f(z)\,\mathrm{d}z = \oint_{C}f(z)\,\mathrm{d}z=2\pi i \sum_{z_0} \mathrm{Res}f(z_0) $$
You can also get the expected results this way.
Another method is to use keyhole contour.
By calculating $\displaystyle{\oint_{C}f(z) \ln z\,\mathrm{d}z}$, you can also get the expected integration result.
Exercise
But note that if the integration path contains singularities, you usually need to change the contour to bypass these singularities. I'll leave a question as an exercise for the OP,
$$ \int_0^{\infty} \frac{\ln x}{x^2-1}\,\mathrm{d}x. $$
You can use the following contour to solve this problem.