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For some time, I have been struggling with the following integral:

$$\int{\frac{x^2+d^2}{\sqrt{\left(x^4+b^2x^2+c^2\right)^3}}\mathrm dx}\;,$$

where $b^4-4c^2>0$.

I did my best, but I am still far away from solving this problem. Does anyone have a clue how to compute it?

I would be very grateful for any hint.

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  • $\begingroup$ It is not elementary, and the general expression is huge... Don't you at least have limits? $\endgroup$ – mickep Jan 6 '15 at 7:40
  • $\begingroup$ Are we to assume you're integrating with respect to $x$? $\endgroup$ – daOnlyBG Jan 6 '15 at 7:42
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    $\begingroup$ @daOnlyBG Personally I'd rather integrate with respect $d$ ;) $\endgroup$ – David H Jan 6 '15 at 7:43
  • $\begingroup$ @DavidH I mean, yeah. Just wanted to know how I'm going to edit this :D $\endgroup$ – daOnlyBG Jan 6 '15 at 7:44
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    $\begingroup$ @M.Vinay I think you're definitely on the right track for how to decompose this monstrosity. I doubt the creator of the problem would have specified the discriminant was positive unless it was to tell us there are real and distinct roots that in principle can be obtained by ordinary algebra. Whether or not that algebra is human-doable in a reasonable amount of time is a different question......... $\endgroup$ – Mathemagician1234 Jan 6 '15 at 7:56
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Hint for first step: This just describes how the initial four parameter problem can simplified to a three parameter one. I find that in solving these types of multi-parameters problems, this should almost always be your first step.

The integral in question may be written as a definite integral with variable lower limit of integration:

$$\mathcal{I}{\left(b,c,d,u\right)}:=\int_{u}^{\infty}\frac{x^2+d^2}{(x^4+b^2x^2+c^2)^{3/2}}\mathrm{d}x,$$

where $b,c,d,u\in\mathrm{R}$ with $b^4-4c^2>0$. For simplicity, we shall assume that the three parameters $b,c,d$ are positive.

Given $c>0$, we can rescale the other three parameters and the integration variable by $\sqrt{c}$, i.e.,

$$\begin{cases} x=y\sqrt{c},\\ \frac{b}{\sqrt{c}}=:\sqrt{2}\,\beta,\\ \frac{d}{\sqrt{c}}=:\delta,\\ \frac{u}{\sqrt{c}}=:\eta,\\ \end{cases}$$

and in the process reduce the original four parameter integral to an integral with one fewer free parameters:

$$\begin{align} \mathcal{I}{\left(b,c,d,u\right)} &=\int_{u}^{\infty}\frac{x^2+d^2}{(x^4+b^2x^2+c^2)^{3/2}}\mathrm{d}x\\ &=\int_{u/\sqrt{c}}^{\infty}\frac{cy^2+d^2}{(c^2y^4+b^2cy^2+c^2)^{3/2}}\sqrt{c}\,\mathrm{d}y\\ &=c^{-3/2}\int_{\eta}^{\infty}\frac{y^2+\delta^2}{(y^4+2\beta^2y^2+1)^{3/2}}\,\mathrm{d}y\\ &=c^{-3/2}\mathcal{I}{\left(\sqrt{2}\,\beta,1,\delta,\eta\right)}.\\ \end{align}$$

That is, it is sufficient to solve,

$$\tilde{\mathcal{I}}{\left(b,d,u\right)}:=\int_{u}^{\infty}\frac{x^2+d^2}{(x^4+2b^2x^2+1)^{3/2}}\,\mathrm{d}x;~~\text{where }b>1.$$


More substantial hint:

Given two real parameters $u,b\in\mathbb{R}$ such that $0\le u$ and $1<b$, define the two-variable function $f{\left(u,b\right)}$ by the definite integral,

$$f{\left(u,b\right)}:=\int_{u}^{\infty}\frac{\mathrm{d}x}{\sqrt{x^4+2b^2x^2+1}}.$$

We'll find it convenient to additionally define the auxiliary parameters,

$$\begin{cases} \varphi:=\arcsin{\left(\frac{1-u^2}{u^2+1}\right)};\\ \kappa:=\sqrt{\frac{b^2-1}{1+b^2}}.\\ \end{cases}$$

Note in particular that, for $0\le u$, we have

$$-1<\frac{1-u^2}{u^2+1}\le 1,$$

and for $1<b$, we have

$$0<\frac{b^2-1}{1+b^2}<1;$$

which guarantee that our parameters $\varphi$ and $\kappa$ are well-defined real numbers.

The integral $f{\left(u,b\right)}$ can be reduced to elliptic integrals of the first kind with two substitutions, $x=\frac{\sqrt{1-y}}{\sqrt{y}}\implies y=\frac{1}{x^2+1}$ and $y=\frac{z+1}{2}\implies z=2y-1$:

$$\begin{align} f{\left(u,b\right)} &=\int_{u}^{\infty}\frac{\mathrm{d}x}{\sqrt{x^4+2b^2x^2+1}}\\ &=\int_{\frac{1}{u^2+1}}^{0}\frac{\left(-\frac{y^{-3/2}}{2\sqrt{1-y}}\right)\,\mathrm{d}y}{\sqrt{\left(\frac{1-y}{y}\right)^2+2b^2\left(\frac{1-y}{y}\right)+1}}\\ &=\frac12\int_{0}^{\frac{1}{u^2+1}}\frac{\mathrm{d}y}{\sqrt{y(1-y)}\sqrt{1+2(b^2-1)y(1-y)}}\\ &=\frac{\sqrt{2}}{2\sqrt{1+b^2}}\int_{-1}^{\frac{1-u^2}{u^2+1}}\frac{\mathrm{d}z}{\sqrt{1-z^2}\sqrt{1-\left(\frac{b^2-1}{1+b^2}\right)z^2}}\\ &=\frac{1}{\sqrt{2}\sqrt{1+b^2}}\int_{-1}^{\sin{\varphi}}\frac{\mathrm{d}z}{\sqrt{\left(1-z^2\right)\left(1-\kappa^2z^2\right)}}\\ &=\small{\frac{1}{\sqrt{2}\sqrt{1+b^2}}\int_{-1}^{0}\frac{\mathrm{d}z}{\sqrt{\left(1-z^2\right)\left(1-\kappa^2z^2\right)}}+\frac{1}{\sqrt{2}\sqrt{1+b^2}}\int_{0}^{\sin{\varphi}}\frac{\mathrm{d}z}{\sqrt{\left(1-z^2\right)\left(1-\kappa^2z^2\right)}}}\\ &=\small{\frac{1}{\sqrt{2}\sqrt{1+b^2}}\int_{0}^{1}\frac{\mathrm{d}z}{\sqrt{\left(1-z^2\right)\left(1-\kappa^2z^2\right)}}+\frac{1}{\sqrt{2}\sqrt{1+b^2}}\int_{0}^{\sin{\varphi}}\frac{\mathrm{d}z}{\sqrt{\left(1-z^2\right)\left(1-\kappa^2z^2\right)}}}\\ &=\frac{K{\left(\kappa\right)}+F{\left(\varphi,\kappa\right)}}{\sqrt{2}\sqrt{1+b^2}}.\\ \end{align}$$

Now try to find a way to express $\mathcal{I}$ as a combination of derivatives of $f$.

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  • $\begingroup$ That's is the first step: 'Scaling'... $\endgroup$ – Felix Marin Jan 6 '15 at 9:28

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