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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.

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    $\begingroup$ Your strategy is$$\int_0^\infty f(x)dx=2i\pi\sum_c\operatorname{Res}_{z=c}f-\lim_{R\to\infty}\int_0^{\pi/2}f(Re^{i\theta})iRe^{i\theta}d\theta-\int_0^\infty if(ix)dx.$$That $f(ix)$ appears limits the $f$ for which this will work. $\endgroup$ Commented Dec 21, 2023 at 23:18
  • $\begingroup$ @J.G., can you pls elaborate it with reference to my example ? Thank you ! $\endgroup$ Commented Dec 22, 2023 at 16:36

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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.

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  • $\begingroup$ Another example, $\displaystyle{\int_0^{1}\frac{x\ln (1+x)}{1+x^4} \, \mathrm{d}x}$. You can solve this integral using the quarter circle contour with a radius of 1. The solution can be found in my blog $\endgroup$ Commented Dec 25, 2023 at 7:29
  • $\begingroup$ Thank you very much for the well detailed answer and explanation, which is exactly I was looking for. If I may ask one question, so the trick is to connect the integrals along the real and imaginary axis, rather than using the ML inequality to show the latter is zero for large enough $|z|$ ? $\endgroup$ Commented Dec 25, 2023 at 23:54
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The answer to your question. Try a function with poles on the imaginary axes, like $\int_0^\infty dx/(x^2+1) = \pi/2$. Since $1/(z^2+1)$ has poles at $z = \pm i$, you can't literally use a quarter circle contour that passes along the imaginary axis to $0$, since (when the contour is big) you'll have a pole along the contour.

It's unrealistic to expect you could always calculate integrals with a certain type of contour when you impose barely any conditions on the integrand (in your case, only asking it to be even). You need to make stronger assumptions. And be careful about thinking you can't calculate an integral by a certain method too: people thought for a long time that the Gaussian integral could not be computed with the residue theorem, but eventually such a method was discovered. See the 1st paragraph in Section 10 here.

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  • $\begingroup$ thank you for your reply but this doesn't answer the question. The singularity for the question that I raise is a double pole at the first quadrant. I know that the contour has to enclose the contour without lying on the contour. $\endgroup$ Commented Dec 22, 2023 at 1:10
  • $\begingroup$ Your question asks "can we always use a large enough quarter circle contour in the 1st quadrant to find the value of the given integral ... when the integrand is an even function?" And my answer provides an example where the quarter circle contour can't be used. The particular example you mentioned was not in the body of the question. $\endgroup$ Commented Dec 22, 2023 at 1:14

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