How Do You Integrate Expressions Like These?

[askor]

How do you you integrate expressions like these?

1. $\frac{1}{\sqrt{x^3 + 3}}$
2. $\frac{1}{\sqrt{x^4 + 3}}$
3. $\frac{1}{x \sqrt{x^5 + 3}}$
4. $\frac{1}{x^2 \sqrt{x^3 + 3}}$
5. $\frac{1}{x^3 \sqrt{x^3 + 3}}$
6. $\frac{1}{x^4 \sqrt{x^3 + 3}}$
7. $\frac{1}{x^5 \sqrt{x^3 + 3}}$

Does trigonometry substitution still eligible?

If not, what method to integrates these expressions?

 

[Mark44]

How do you you integrate expressions like these?

1. $\frac{1}{\sqrt{x^3 + 3}}$
2. $\frac{1}{\sqrt{x^4 + 3}}$
3. $\frac{1}{x \sqrt{x^5 + 3}}$
4. $\frac{1}{x^2 \sqrt{x^3 + 3}}$
5. $\frac{1}{x^3 \sqrt{x^3 + 3}}$
6. $\frac{1}{x^4 \sqrt{x^3 + 3}}$
7. $\frac{1}{x^5 \sqrt{x^3 + 3}}$

Does trigonometry substitution still eligible?

If not, what method to integrates these expressions?

I don't think trig substitution could be used in any of them. If #2 were $\frac x {x^4 + 3}$ instead of what you wrote, then it would be susceptible to a trig substitution. What you have an integral that involves a square root of a sum or difference of squares, then trig substitution is a reasonable choice of method. Otherwise the integrals get pretty tricky and might not have nice, closed form solutions.

 

[PeroK]

How do you you integrate expressions like these?

1. $\frac{1}{\sqrt{x^3 + 3}}$
2. $\frac{1}{\sqrt{x^4 + 3}}$
3. $\frac{1}{x \sqrt{x^5 + 3}}$
4. $\frac{1}{x^2 \sqrt{x^3 + 3}}$
5. $\frac{1}{x^3 \sqrt{x^3 + 3}}$
6. $\frac{1}{x^4 \sqrt{x^3 + 3}}$
7. $\frac{1}{x^5 \sqrt{x^3 + 3}}$

Does trigonometry substitution still eligible?

If not, what method to integrates these expressions?

See what Wolfram Alpha thinks.

 

[mathwonk]

Not everything can be integrated, at least not in the usual sense of finding a familiar function whose derivative equals your given function. I.e. not every integral can be integrated in "elementary terms". You are asking for an antiderivative, or function whose derivative equals your function. But you don't know most functions, and there are a lot of functions you don't know, have innocent looking derivatives that you do know, like the function in 1. above. I.e. the derivative of a familiar function is still familiar, but there may be very unfamiliar functions whose derivatives look very familiar, so just because you have a familiar looking function, there is no reason to expect that you have ever seen its antiderivative. Indeed this is a famous way to discover or create new interesting functions, by defining them as the antiderivative of some continuous function whose anitiderivative you have not met. Just as the exponential function can be defined as the inverse of the integral of 1/x, and the sin can be defined as the inverse of the integral of 1/sqrt(1-x^2), so can more exotic functions be defined as the inverse of other rational integrals.

The integral of 1. is the inverse of what is called an "elliptic function". This is a doubly periodic function, hence somewhat of a generalization of the sin function, which is simply periodic. The level of complexity of the function defined by (inverse of) the integral of 1/sqrt(p(x)), where p is a polynomial, is measured by the topological genus of the plane curve y^2 = p(x). Hence when p is a cubic, the genus is one and this is an "elliptic curve". This is a famous classical story and leads to the theory of Riemann surfaces. A wonderful reference is the 3 volume treatment by Carl Siegel, Topics in complex function theory.

 

[lurflurf]

It is natural to substitute the contents of the square root.
$$u=x^b+3$$ In 3 and 6 we can finish the integral easily. The others we can use elliptic integrals as mathwonk recommends or hypergeometric functions.
For example use
$$\frac{1}{x^{a+1}\sqrt{x^b+3}}=
-\frac{1}{a\sqrt{3}}\dfrac{d}{dx}\frac{1}{x^a}
\,
_2F_1\left(\frac{1}{2},-\frac
{a}{b};1-\frac{a}{b};-\frac{x
^b}{3}\right)
$$

 

[mathwonk]

One can study functions z defined by integrals like dx/sqrt(1-x^2) in terms of the differential equations they satisfy. By the FTC, dz/dx = 1/sqrt(1-x^2), so if f(z) = x(z) is the inverse function, then x(z) satisfies the d.e. f’ = dx/dz = sqrt(1-x^2), or (f’)^2 = 1-f^2. So if you know or can find a solution to this differential equation, then its inverse function is your original integral. Of course cos and sin do satisfy this d.e.

Riemann took the view of trying to understand the qualitative behavior of functions defined by integrals like that of dx/sqrt(1-x^2), or dx/(sqrt(x^3+3), or more general ones, in the most natural setting. In particular he tried to deal with the problem of multi valued- ness as far as possible. First of all, the integrand, or differential, such dx/sqrt(1-x2), is not well defined, or rather it is multi valued, since there are two values of the square root.

Thus Riemann chose to view this differential as defined, not on the x axis, where it is 2 valued, but on the plane curve y^2 = 1-x^2, where it is the single valued function y. I.e. this plane curve has 2 points over most points of the x axis, and y takes the 2 different values of the square root at these 2 points. Thus by doubling the size of the domain, a 2- valued function has become single valued. Now if z is the function defined as the integral z = integral of dx/sqrt(1-x^2), evaluated from a base point x0 to a variable point x, The function z is 2 - valued on an interval of the x-axis abiding x=1 and x=-1, but we now view it as the function z = integral of dx/y, along a path from a base point (x0,y0) to the variable point (x,y), lying on the plane curve M: {y^2 = 1-x^2}. Now it is even more multiple valued since there are many paths from one point to another, but it is at least defined on the whole circle, (and it is single valued on the set of paths from the base point).

Now the original function on (intervals of) the x axis, defined as an integral, has been decomposed into a composition of two functions, where one is the function from the x-axis to the plane curve M, defined by x—>(x, sqrt(1-x^2)), which of course is only defined on a small subset of the x axis, avoiding the points x = 1, -1, (and depends on a choice of square root), and then composed with the function defined on the curve M as the integral of dx/y. Notice that where y=0, dx also equals zero, since the projection from the circle to the x-axis is ramified where y=0, hence the quotient dx/y is still analytic where y=0.

Now the first function is just the implicitly defined function y(x), and can be studied e.g. by power series expansions if desired. The more interesting part then for us is the second part defined by the integral of dx/y, which is defined on the plane curve by integrating along a path from a base point (x0,y0) to a variable point (x,y) on the curve. This function depends on the choice of path, and thus is also multi - valued as a function on the plane curve y^2 = 1-x^2, which is just the unit circle. If we assume then we understand the function x—>(x,sqrt(1-x^2)), our original integral function is determined just by the integral of dx/y on the unit circle.

This turns out to be the arc length function, and is multi valued since the arc we choose from our base point to our variable point may wind several times around the circle before stopping at the variable point. The composition from a point x on the x axis, up (or down) to a point on the unit circle, followed by arc length from a base point to that point, is just the arcsin function. Although this arcsin function from the circle to the real line is multiple valued, we can force it to be single valued by identifying the different values on the real line taken by the function. Then we get a single valued function from the circle to the quotient space R/2πZ, i.e. we identify two arc lengths if they differ by an integer multiple of 2π. E.g. the arc length from the base point to itself is any element of the set {…, -2π, 0, 2π, 4π,…} (Thus we have seen two tricks for changing multi valued functions into single valued ones, enlarging the domain, and collapsing the range.)

Then the inverse function f(z) = sin(z), satisfies the differential equation (f’)^2 = 1-f^2, since sin’ = cos, and cos^2 = 1-sin^2, so we have a parametrization “by arc length” of the unit circle sending z to (f(z),f’(z)) = (sin(z), cos(z)). (I guess I could have chosen a non standard branch of the inverse trig functions so that I would have had the solution as f = arccos, so this would have given the usual parametrization z—>(cos(z), sin(z)).) I apologize for that. Either way, this is the universal covering map of the unit circle by the real line.

We can follow this analysis also for the integral of the multi valued differential dx/sqrt(x^3+3) on the x axis. We consider this first as the single valued differential dx/y, defined on the plane curve M: {y^2 = x^3+3}, and focus attention on the integral of the single valued differential dx/y, as a multivalued function on M, by integrating along paths on M starting at a base point (x0,y0), up to a variable point (x,y). This time, this differential remains analytic even at the points “at infinity” of this plane curve, which we view in the complex projective plane, where it is topologically a torus, or elliptic curve.

So this integral defines a multi valued function g from the torus to the complex numbers, analogous to the arc length function from the circle to the real line. This time of course the torus has two holes or loops on it, and going around either loop with our path, adds a constant “period” to the value of our integral. In an analogous way to the sin function, the inverse function f is thus a doubly periodic function on the complex numbers, which satisfies the differential equation (f’)^2 = f^3 + 3, hence the mapping z—>(f(z), f’(z)) gives a parametrization of the torus, (the projective completion of y^2 = x^3 +3), and coincides with the universal covering map of the torus. (The Weierstrass P-function is an example of such an f.)

Without taking inverses, the function g itself can again be made single valued by modding out the complex numbers by the lattice spanned by the two (complex) constant period vectors a and b, of the differential, integrated around the two loops in the torus, giving a well defined map M—>C/{Za+Zb}, which turns out to be an isomorphism.

In his search for general principles, Riemann asked himself not just which curve does the differential dx/sqrt(x^3+3) live on in the plane, but what are the other differentials that live on this curve? In particular he proved that up to constant multiple, this is the only everywhere analytic differential on the associated torus, i.e. that the vector space of such differentials has dimension one. He proved also that this is because the torus has topological genus one. He took this approach because he did not want to just explore each function he came upon one at a time, but to organize them according to some principle, and the organizing principle is whether they live on the same complex curve, (i.e. a real topological surface with complex coordinates).

In the case of the differential dx/sqrt(1-x^5), the plane curve y^2 = 1-x^5 has singular projective completion, but the desingularized curve M has topological genus 2, and hence has 2 diml space of analytic differentials. Namely in addition to dx/y, xdx/y is also analytic. Hence if their integrals are the multi valued functions z and w, the inverse of either one of them will satisfy the differential equation (f’)^2 = 1-f^5, and hence give a parametrization of the plane curve M.

Since there is no way to prefer one of these differentials to the other, to understand the curve M, it is more natural to consider the (multi valued) map M—> CxC, to the product of 2 copies of the complex numbers, sending t—->(z(t),w(t)). Since M has genus 2, there are 4 independent “homology” loops on it, so there are 4 complex constant periods that are added to a path integral each time it goes around one of these loops. If these period vectors are a,b,c,d, we get a single valued mapping M—>CxC/{Za+Zb+Zc+Zd}, which embeds the surface M in the 4 dimensional “complex 2- torus” CxC/{Za+Zb+Zc+Zd}, which is topologically like real 4 space modded out by the rank 4 lattice spanned by the integer vectors on the 4 axes. Riemann also wrote down an analytic "theta function" which in this case cuts out exactly the image of the genus 2 curve in the complex torus, its "Jacobian" variety.

This embedding of M in its Jacobian is the famous Abel map, defined by Abelian integrals, and it generalizes to all complex curves of every genus, embedding a curve of genus g, into a g complex dimensional torus. Then analytic functions on the torus, like theta functions, can be restricted to the curve and their properties studied. Special curves of the sort studied here, of form y^2 = p(x), where p is a polynomial of degree 2g+1, or 2g+2, with distinct roots, have genus g, and are called "hyperelliptic", since like elliptic curves they are double covers of the x axis.

In the first case of the sin function, the corresponding complex curve whose real points are on the circle, is a topological sphere of genus zero, which thus has no non zero analytic differentials at all. The puzzle that dx/y seems to give one, being analytic where y=0, is explained by the fact that this differential has a simple pole at each of the two points "at infinity" in the projective complex plane. In general the number of zeroes minus the number of poles of a differential equals 2g-2, where g is the genus.

Recall that when Riemann made this study, the topology of surfaces did not exist, so he invented it, including its homology, for this purpose in his dissertation and the follow up paper on Abelian functions. These are highly recommended of course, but famously challenging to read. Those parts I read took me about a day per page. (My brief review of the English translation of Riemann's works is post #112 here
https://www.physicsforums.com/threads/best-all-time-mathematicians-physicists.53558/page-4
There is also the "expository" "Riemann's theory of algebraic functions and their integrals" by Felix Klein, which physicists will probably be able to appreciate more than I was.

Forgive me for the long post, I got fascinated by this question, and needed to think about it carefully before posting an answer.

By the way, looking at the examples above in the light of topology, they do not appear to be as complicated as they seem at first. I.e. most of the polynomials involved have multiple roots, e.g. in #7: y^2 = x^10(x^3+3), and this lowers the genus of the "desingularized" surface below what the degree formula would give. E.g. all except #3 seem to have genus one, and #3 seems to have genus 2. I could be wrong as I have just drawn a picture and not done the calculations.

 

[Mayhem]

I have not evaluated these personally, but I recall solving similar problems where completing the square in-place allowed for analytical solutions. Again, might not be applicable in this case.

Edit: also clever substitutions go a very long way. Sometimes the substitution requires days of thinking to see.