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The indefinite integral below, quoted from Gradshteĭn's Table of Integrals, Series, and Products, 7th ed., (bottom of p.104) appears to contain at least two typos (highlighted in purple).

$\mathbf{2.291.4}$ $$\begin{align} &\int\frac{dx}{\sqrt{a+bx+cx^2+dx^3+cx^4+bx^5+ax^6}}\\ &~~~~~~=-\frac{1}{\sqrt{2}}\int\frac{\color{purple}{dx}}{\sqrt{(z+1)p}}-\frac{1}{\sqrt{2}}\int\frac{dz}{\sqrt{(z-1)p}},~~\left[x=z+\sqrt{z^2-1}\right]\\ &~~~~~~=-\frac{1}{\sqrt{2}}\int\frac{\color{purple}{d}}{\sqrt{(z+1)p}}+\frac{1}{\sqrt{2}}\int\frac{dz}{\sqrt{(z-1)p}},~~\left[x=z-\sqrt{z^2-1}\right] \end{align}$$ where $p=2a(4z^3-3z)+2b(2z^2-1)+2cz+d$. $~~~\square$

The second typo is obviously typographical, but the first typo is substantial if I'm correct. I believe in both cases the highlighted portions should read "$dz$". I tried confirming the typo by deriving the correct formula firsthand, but I quickly ran into difficulties and decided I needed help.



Edit

I feel confident now in presuming that both highlighted portions should instead read $dz$. My thanks to mike for providing some insightful observations that removed pretty much any doubt I had of this. That said I, I'd very much like to see a complete derivation before signing off from this question, all the more so now that it seems almost within reach. At this point my primary interest here is less about proving this integral correct than it is about spotting techniques that can be generalized or re-purposed towards similar integrals in the future.

And one final request. If anyone knows any good references on pseudo-elliptic integrals, I would be very grateful if you could point them out to me. Cheers.

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  • $\begingroup$ I think that both $d$ and $dx$ should be $dz$. $\endgroup$ – mike Sep 7 '14 at 21:31
  • $\begingroup$ It can be concluded by dimensional analysis that the lone $d$ is supposed to be a $dx$ or $dz$. $\endgroup$ – user111187 Sep 7 '14 at 21:33
  • $\begingroup$ another way is to verify them numerically with some definite numerical integration $\endgroup$ – mike Sep 7 '14 at 21:36
  • $\begingroup$ @mike I think they're both supposed to be $dz$ too. But the presence of multiple typos in a single statement that I can't independently corroborate make me anxious other problems might exist. $\endgroup$ – David H Sep 7 '14 at 21:43
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    $\begingroup$ I checked the trivial special case $d=1$, $a=b=c=0$. The formulas with $dz$ work. As for corroboration in the general case, I'd put $z=(1+x^2)/(2x)$ in the given formulas, thus eliminating $z$ from everywhere. Then differentiate in $x$ (in effect, removing the integral signs) and compare. With CAS, of course... $\endgroup$ – user147263 Sep 7 '14 at 21:51
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here is what I got towards verification of the problem.

Define $$Q(x)=a+bx+cx^2+dx^3+cx^4+bx^5+ax^6\tag{0}$$

Substitution of $$x=z-\sqrt{z^2-1}\tag{1}$$

into $x^3 Q(x)$ leads to:

$$x^3 Q(x)=p(z) \tag{2}$$ $$p= 2a( 4z^3-3z)+2b(2z^2-1) + 2cz +d\tag{3}$$

From (2.291.4.2) we have: $$\int\frac{dx}{\sqrt{Q(x)}} =-\frac{1}{\sqrt{2}}\int\frac{dz}{\sqrt{(z+1)p}}+\frac{1}{\sqrt{2}}\int\frac{dz}{\sqrt{(z-1)p}}\tag{4}$$

Using (3), we know that we need to prove:

$$\sqrt{x^3}dx =-\frac{1}{\sqrt{2}}\frac{dz}{\sqrt{z+1}}+\frac{1}{\sqrt{2}}\frac{dz}{\sqrt{z-1}}\tag{5}$$

or

$$\sqrt{x^3}dx =-\frac{1}{\sqrt{2}}\frac{d(z+1)}{\sqrt{z+1}}+\frac{1}{\sqrt{2}}\frac{d(z-1)}{\sqrt{z-1}}\tag{6}$$

to be continued...

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  • $\begingroup$ +1) Great observation. I used WRA to substitute $x=z-\sqrt{z^2-1}$ into $Q(x)$ directly and got back $Q(x)=p(z)\cdot \varphi(z)$, where $\varphi$ was a complicated algebraic function of $z$ which I couldn't make sense of. Recognizing that it represents $x^{-3}$ clears a number of things up already. I do still want to see a complete derivation though, if you see a way to do it. $\endgroup$ – David H Sep 7 '14 at 22:21
  • $\begingroup$ I am trying. What we need to figure is figure out $\frac{dx}{\sqrt{Q(x)}}=A \frac{dz}{\sqrt{(z+1)p(z)}}+B \frac{dz}{\sqrt{(z-1)p(z)}}$ $\endgroup$ – mike Sep 7 '14 at 22:28
  • $\begingroup$ In your line $(4)$, you mean to have $dx=\left(1-\frac{z}{\sqrt{z^2-1}}\right)dz$, right? $\endgroup$ – David H Sep 7 '14 at 22:33
  • $\begingroup$ yes. but I also got another simpler one as shown above. $\endgroup$ – mike Sep 7 '14 at 22:46
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For convenience, I borrow mike's notation and define,

$$Q(x):=a+bx+cx^2+dx^3+cx^4+bx^5+ax^6.$$

As indicated in the problem statement above, I also define,

$$p(z):=p=2a(4z^3-3z)+2b(2z^2-1)+2cz+d.$$

I have had a considerable amount of trouble trying to transform the integral over $x$ into the desired integral over $z$ via the substitutions $x=z\pm\sqrt{z^2-1}$, i.e., trying to show directly that:

$$\int\frac{\mathrm{d}x}{\sqrt{Q(x)}}=-\frac{1}{\sqrt{2}}\int\frac{\mathrm{d}z}{\sqrt{(z+1)\,p(z)}}\mp\frac{1}{\sqrt{2}}\int\frac{\mathrm{d}z}{\sqrt{(z-1)\,p(z)}};~\left[x=z\pm\sqrt{z^2-1}\right].$$

As mike explained in his response, the key step is demonstrating that

$$\sqrt{x^3}dx =-\frac{1}{\sqrt{2}}\frac{dz}{\sqrt{z+1}}\mp\frac{1}{\sqrt{2}}\frac{dz}{\sqrt{z-1}},$$

but despite all my efforts I've been unsuccessful at proving the above relation.

It was suggested in a comment (now removed) that we work backwards and instead transform the integral over $z$ into the integral over $x$. Surprisingly (to me at least), the transformation turns out to be much clearer in reverse. The inverse relation for the substitution formulas

$$x=\varphi_{\pm}{(z)}=z\pm\sqrt{z^2-1}$$

is,

$$z=\varphi_{\pm}^{-1}{(x)}=\frac{x^2+1}{2x}.$$

Note that the domain of both of the functions $\varphi_{\pm}{(z)}$ is $(-\infty,-1]\cup[1,+\infty)$, whereas the range of $\varphi_{+}{(z)}$ is $[-1,0)\cup[1,+\infty)$ while the range of $\varphi_{-}{(z)}$ is $(-\infty,-1]\cup(0,1]$.

The differential $\mathrm{d}z$ is transformed under the substitution $z=\frac{x^2+1}{2x}=\frac12\left(x+\frac{1}{x}\right)$ to:

$$\mathrm{d}z=\frac{x^2-1}{2x^2}\,\mathrm{d}x.$$

We also have:

$$\begin{align} p(z) &=2a(4z^3-3z)+2b(2z^2-1)+2cz+d\\ &=a\frac{x^6+1}{x^3}+b\frac{x^4+1}{x^2}+c\frac{x^2+1}{x}+d\\ &=\frac{Q(x)}{x^3};\\ \end{align}$$

$$z+1=\frac{(x+1)^2}{2x};$$

$$z-1=\frac{(x-1)^2}{2x}.$$

And so, the integrals over $z$ transform as:

$$\begin{align} \frac{1}{\sqrt{2}}\int\frac{\mathrm{d}z}{\sqrt{(z\pm1)\,p(z)}} &=\frac{1}{\sqrt{2}}\int\frac{\left(\frac{x^2-1}{2x^2}\right)\,\mathrm{d}x}{\sqrt{\frac{(x\pm1)^2}{2x}\cdot\frac{Q(x)}{x^3}}}\\ &=\frac12\int\frac{\left(x^2-1\right)\,\mathrm{d}x}{\left|x\pm1\right|\sqrt{Q(x)}}\\ &=\frac12\int\operatorname{sgn}{\left(x\pm1\right)}\frac{\left(x\mp1\right)\,\mathrm{d}x}{\sqrt{Q(x)}}\\ \end{align}$$

The above integrals probably shed some light on what was causing the difficulty beforehand. Because of the $\text{sgn}$-function factors, the only way to reduce the sum/differences of the integrals to the desired form $\int\frac{\mathrm{d}x}{\sqrt{Q(x)}}$ is by restricting the variable $x$ to an appropriate interval. We can easily check that,

$$ 1= \begin{cases} \frac12(x-1)\operatorname{sgn}{\left(x+1\right)}-\frac12(x+1)\operatorname{sgn}{\left(x-1\right)};~~x<-1\\ -\frac12(x-1)\operatorname{sgn}{\left(x+1\right)}-\frac12(x+1)\operatorname{sgn}{\left(x-1\right)};~~-1<x<1\\ -\frac12(x-1)\operatorname{sgn}{\left(x+1\right)}+\frac12(x+1)\operatorname{sgn}{\left(x-1\right)};~~x>1.\\ \end{cases}$$

As such, it may make the casework clearer to consider definite integrals instead of indefinite ones.

Let's assume an integration interval of $[x_1,x_2]\subseteq(0,1]\subseteq\text{range }{\varphi_{-}}$. Then,

$$\begin{align} \int_{x_{1}}^{x_{2}}\frac{\mathrm{d}x}{\sqrt{Q(x)}} &=\int_{x_{1}}^{x_{2}}\frac{\left[-\frac12(x-1)\operatorname{sgn}{\left(x+1\right)}-\frac12(x+1)\operatorname{sgn}{\left(x-1\right)}\right]\,\mathrm{d}x}{\sqrt{Q(x)}}\\ &=-\frac12\int_{x_{1}}^{x_{2}}\operatorname{sgn}{\left(x+1\right)}\frac{\left(x-1\right)\,\mathrm{d}x}{\sqrt{Q(x)}}-\frac12\int_{x_{1}}^{x_{2}}\operatorname{sgn}{\left(x-1\right)}\frac{\left(x+1\right)\,\mathrm{d}x}{\sqrt{Q(x)}}\\ &=-\frac{1}{\sqrt{2}}\int_{\varphi_{-}^{-1}{(x_{1})}}^{\varphi_{-}^{-1}{(x_{2})}}\frac{\mathrm{d}z}{\sqrt{(z+1)\,p(z)}}-\frac{1}{\sqrt{2}}\int_{\varphi_{-}^{-1}{(x_{1})}}^{\varphi_{-}^{-1}{(x_{2})}}\frac{\mathrm{d}z}{\sqrt{(z-1)\,p(z)}},\\ \end{align}$$

which is the expected formula. We would also also arrive at the (corrected) formula given in Gradshteyn had the integration interval instead been $[x_1,x_2]\subseteq[1,\infty)\subseteq\text{range }{\varphi_{+}}$. However, it appears that Gradshteyn's formulas fail from sign errors when $x<0$. It is my belief that this missing condition in the text's statement of the proposition explains my initial trouble with the forward derivation.

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