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I have the following principal value integral: $$\mathcal{P}\int_0^\infty\frac{x^4}{\left(1+\frac{x^2}{B^2}\right)^{4}\sqrt{C^2+x^2}\left[\sqrt{C^2+a^2}-\sqrt{C^2+x^2}\right]}dx$$ where $a,B,C\in\mathbb{R}^+$, (and of course the singularity is at $x=a$). If it's analytically solvable (which I highly doubt), I'd like to know how to solve it. (As a commenter has pointed out, there is actually an elementary antiderivative for the indefinite integral, but the form is rather complicated so I'd rather just reduce it to a non-principal form and evaluate it numerically.) How can I manipulate it to equate it to an expression in terms of "ordinary" (non-principal-value) integrals?

It has been suggested by someone I know that I might be able to make use of the result: $$\mathcal{P}\int_0^\infty\frac{dx}{C^2-x^2}=0$$ for $C\in\mathbb{R}$, but I am unsure how to make this result useful.

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    $\begingroup$ The indefinite integral, in fact, has elementary antiderivative, Use the substitution $$t + x= \sqrt{C^2+x^2} $$ (known as Euler substitution) to convert it into rational function. $\endgroup$ Commented Jan 25, 2018 at 16:09
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    $\begingroup$ If your purpose is to numerically calculate it for various $a,B,C$, then since the integrand is even, you can shift the line of integration to $\pm \infty + i\varepsilon$, but you have to include half the residue at $x=a$. $\endgroup$ Commented Jan 25, 2018 at 16:16
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    $\begingroup$ I haven't really studied any complex analysis, so I'm unsure what you mean by the second comment. As for the first comment - seems doable, but doesn't the result still have a singularity? $\endgroup$ Commented Jan 26, 2018 at 4:32

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Chop out the singularity piece by piece: $$\begin{align}f(x,a,B,C)&=\frac{x^4}{\left(1+\frac{x^2}{B^2}\right)^4\sqrt{C^2+x^2}\left(\sqrt{C^2+a^2}-\sqrt{C^2+x^2}\right)}\\ &=\frac{x^4\left(\sqrt{C^2+a^2}+\sqrt{C^2+x^2}\right)}{\left(1+\frac{x^2}{B^2}\right)^4\sqrt{C^2+x^2}\left(a^2-x^2\right)}\\ &=\frac{x^4}{\left(1+\frac{x^2}{B^2}\right)^4\left(a^2-x^2\right)}\left[\frac{\sqrt{C^2+a^2}+\sqrt{C^2+x^2}}{\sqrt{C^2+x^2}}-2+2\right]\\ &=\frac{x^4}{\left(1+\frac{x^2}{B^2}\right)^4\left(a^2-x^2\right)}\left[\frac{\sqrt{C^2+a^2}-\sqrt{C^2+x^2}}{\sqrt{C^2+x^2}}+2\right]\\ &=\frac{x^4}{\left(1+\frac{x^2}{B^2}\right)^4\left(a^2-x^2\right)}\left[\frac{a^2-x^2}{\sqrt{C^2+x^2}\left(\sqrt{C^2+a^2}+\sqrt{C^2+x^2}\right)}+2\right]\\ &=\frac{x^4}{\left(1+\frac{x^2}{B^2}\right)^4\sqrt{C^2+x^2}\left(\sqrt{C^2+a^2}+\sqrt{C^2+x^2}\right)}+\frac{2x^4}{\left(1+\frac{x^2}{B^2}\right)^4\left(a^2-x^2\right)}\end{align}$$ Then $$\begin{align}\frac{2x^4}{\left(1+\frac{x^2}{B^2}\right)^4\left(a^2-x^2\right)}&=\frac1{\left(1+\frac{x^2}{B^2}\right)^4\left(a^2-x^2\right)}\left[2x^4-2a^4+2a^4\right]\\ &=\frac1{\left(1+\frac{x^2}{B^2}\right)^4\left(a^2-x^2\right)}\left[2\left(a^2+x^2\right)\left(a^2-x^2\right)+2a^4\right]\\ &=\frac{2\left(a^2+x^2\right)}{\left(1+\frac{x^2}{B^2}\right)^4}+\frac{2a^4}{\left(1+\frac{x^2}{B^2}\right)^4\left(a^2-x^2\right)}\end{align}$$ And finally $$\begin{align}\frac{2a^4}{\left(1+\frac{x^2}{B^2}\right)^4\left(a^2-x^2\right)}&=\frac{2a^4}{\left(a^2-x^2\right)}\left[\frac1{\left(1+\frac{x^2}{B^2}\right)^4}-\frac1{\left(1+\frac{a^2}{B^2}\right)^4}+\frac1{\left(1+\frac{a^2}{B^2}\right)^4}\right]\\ &=\frac{2a^4}{\left(a^2-x^2\right)}\left[\frac{\left(\frac4{B^2}+\frac6{B^4}\left(a^2+x^2\right)+\frac4{B^6}\left(a^4+a^2x^2+x^4\right)+\frac1{B^8}\left(a^6+a^4x^2+a^2x^4+x^6\right)\right)\left(a^2-x^2\right)}{\left(1+\frac{x^2}{B^2}\right)^4\left(1+\frac{a^2}{B^2}\right)^4}+\frac1{\left(1+\frac{a^2}{B^2}\right)^4}\right]\\ &=\frac{2a^4\left(\frac4{B^2}+\frac6{B^4}\left(a^2+x^2\right)+\frac4{B^6}\left(a^4+a^2x^2+x^4\right)+\frac1{B^8}\left(a^6+a^4x^2+a^2x^4+x^6\right)\right)}{\left(1+\frac{x^2}{B^2}\right)^4\left(1+\frac{a^2}{B^2}\right)^4}\\ &+\frac{2a^4}{\left(1+\frac{a^2}{B^2}\right)^4\left(a^2-x^2\right)}\end{align}$$ So that last term embodies the singularity with a principal value of $0$ and hopefully you can integrate the rest of it numerically.

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