反三角函数（2）

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Solution
What comes to mind is to maybe write the integrand as:

Now, deal with

I think ben‘s idea of contours may be a cool way to go in this case.
Note the pole at \(\sqrt{b}i\) and \(\tan^{-1}(x^{2})\) has singularities at \(\pm \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i\)

EDIT:
Please check my reasoning and feel free to add your part.
I looked at this and came up with something rough. I need to tweak it. The branch of arctan and all that gets me sideways. But, something looks to be coming together.
Please, feel free to finish the idea.

Using the expansion at (1), the residues at the arctan singularity at \(\pm \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i\) is 0.
With the residue at \(z=\sqrt{2-\sqrt{3}}i\) we then have:

with the residue at \(z=\sqrt{2+\sqrt{3}}i\), we have:

subtract per the expansion at the top and arrive at:

Now, it appears we have to divide by 2 to arrive at the final answer. I believe is due to the symmetry of the integrand:
That is, something like

反三角函数（2）

标签：hat   splay   nal   note   mat   free   finish   div   appear   

原文地址：https://www.cnblogs.com/chugou12438/p/14136125.html