I am trying to solve (by hand) the integral:$$\int_0^∞\frac1x\left(\frac{(x+a)^2}{((x+a)^2+b)^{3/2}}-\frac{(x-a)^2}{((x-a)^2+b)^{3/2}}\right)\,\mathrm dx,$$

where $a$ and $b$ are real and positive. Mathematica actually generates the answer, but I am stumped on how to solve it by hand and quite curious. Answer:


Is there a (reasonably) nice way to solve something like this? Any tips on how to proceed?

  • $\begingroup$ i think you could go with hyperbolic trigometic substituition but im not 100% sure. $\endgroup$ – cand Mar 13 at 5:55

Let $I$ denote the integral. By a simple substitution, we find that

$$ I = \mathrm{PV}\!\int_{-\infty}^{\infty} \frac{1}{x-a} \frac{x^2}{(x^2+b)^{3/2}} \, \mathrm{d}x = \operatorname{Re}\left( \int_{-\infty + i\epsilon}^{\infty+i\epsilon} \frac{1}{z-a} \frac{z^2}{(z^2+b)^{3/2}} \, \mathrm{d}z \right), $$

where the path of integration in the last step denotes any straight line $\operatorname{Im}(z)=\epsilon \in (0, \sqrt{b})$ oriented to the right, and the principal branch cut is adopted for $(z^2+b)^{3/2}$. Taking integration by parts,

$$ I = - \operatorname{Re}\left( \int_{-\infty + i\epsilon}^{\infty+i\epsilon} \frac{a}{(z-a)^2} \frac{1}{\sqrt{z^2+b}} \, \mathrm{d}z \right). $$

We notice that its branch-cut in the upper-half plane is $i[\sqrt{b},\infty)$. So, by deforming the contour to wrap this branch-cut, we have

$$ I = \operatorname{Re}\left( \int_{\sqrt{b}}^{\infty} \frac{2a}{(y+ia)^2\sqrt{y^2-b}} \, \mathrm{d}y \right) = \int_{\sqrt{b}}^{\infty} \frac{2a(y^2-a^2)}{(y^2+a^2)^2\sqrt{y^2-b}} \, \mathrm{d}y. $$

As for further simplification, substitute $p = b/a^2$ and $y = \sqrt{\frac{b}{1-s}}$. Then

\begin{align*} I &= \frac{1}{a} \int_{0}^{1} \frac{p-1+s}{(p+1-s)^2\sqrt{s}} \, \mathrm{d}s \\ &= \frac{1}{a} \left[ \frac{2p\sqrt{s}}{(p+1)(p+1-s)} - \frac{2}{(p+1)^{3/2}}\operatorname{arctanh}\left(\sqrt{\frac{s}{p+1}}\right) \right]_{0}^{1} \\ &= \frac{2}{a} \frac{\sqrt{p+1} - \operatorname{arctanh}\left(\sqrt{\frac{1}{p+1}}\right)}{(p+1)^{3/2}}. \end{align*}

Plugging $p = b/a^2$ back shows that this expression coincides with the answer provided in OP, and so, the desired equality is proved.

  • $\begingroup$ Thanks! How did you figure out the substitution $y=\sqrt{\frac{b}{1-s}}$? With everything else I can see where the intuition came from, but that step was too perfect. Is there a trick for spotting that? $\endgroup$ – WillG Mar 14 at 3:32
  • $\begingroup$ @WillG, I admit that the substitution is not an easy one to come up with, and in fact, I had to apply chain of substitutions $y=\sqrt{b}\,u$, $u=1/\sqrt{t}$, and $t=1-s$ to reach this. $\endgroup$ – Sangchul Lee Mar 14 at 3:56
  • 1
    $\begingroup$ @SangchulLee Your chain of substitutions seem to me like they might just be an Euler substitution in disguise. That would of course be easiest way to spot such a substitution from the outset instead backing into it through a sequence of simpler substitutions. $\endgroup$ – David H Mar 17 at 18:14

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