How to compute this integral ?

y33t

As the title obviously states, how can I evaluate this integral by hand ? I know the result of it, I need to learn how to do it.

∫(z²+x²)^-3/2dx

Thanks in advance.

arildno

Letting "x" be the variable to integrate first, set x=z*Sinh(u), where Sinh(u) is the hyperbolic sine function.

y33t

I didn't get it, it's from a simple example from a textbook. There should be a simpler way of doing this because it just skips the evaluation of this integral and directly passes to the result as if it's "that" easy to do. Any simpler solutions ?

arildno

No.

it is NOT a particularly easy integral to evaluate, in that it is rather lengthy to do so.
That is probably why your book skipped it.

dextercioby

x=z tan t can also do the trick.

arildno

Sure enough.
Scribbling out the solution with tan(t) or Sinh(t) takes about the same amount of time and space, though..

arildno

The actual derivation is rather lengthy, but here it is:
1. x=zSinh(t).
Thus, we have:
[tex]dx=z\cosh(t)dt[/tex]
[tex]z^{2}+x^{2}=z^{2}(1+\sinh^{2}(t))=z^{2}\cosh^{2}(t)[/tex]
Thus, the integral can be simplified to:
[tex]\int\frac{dt}{z|z|\cosh^{2}(t)}[/tex]
2. This is readily integrated to:
[tex]\frac{\tanh(t)}{z|z|}=\frac{x}{z^{2}\sqrt{z^{2}+x^{2}}}[/tex]

HallsofIvy

Arildno really likes the hyperbolic function substitutions. Personally, I prefer trig substitutions, perhaps only because they were the first ones I learned.

We know, of course, that [itex]sin^2(\theta)+ cos^2(\theta)= 1[/itex] and, dividing through by [itex]cos^2(\theta)[/itex], [itex]tan^2(\theta)+ 1= sec^2(\theta)[/itex].

So if we let [itex]x= z tan(\theta)[/itex], [itex]z^2+ x^2= z^2+ z^2tan^2(\theta)= z^2(1+ tan^2(\theta)_= z^2sec^2(\theta)[/itex]. Of course, [itex]dx= z sec^2(\theta)d\theta[/itex] so the integral becomes
[tex]\int\frac{z sec^2(\theta)}{z^3 sec^3(\theta)}d\theta= \int \frac{1}{sec(\theta)}d\theta[/tex]
[tex]= \int cos(\theta) d\theta[/tex]
which is easy.

Since [itex]\theta= arctan(x/z)[/itex], the integral will eventually give [itex]sin(arctan(x/z))[/itex]. You can imagine that as describing a right triangle with legs x and z (x opposite the angle) so that the hypotenuse has length [itex]\sqrt{x^2+ z^2}[/itex] and [itex]sin(arctan(x/z))= \frac{x}{\sqrt{x^2+ z^2}}[/itex].

swle

Sorry that this is a bit late, but is Z a real number?
If not then you could use Cauchy: non-holomorphic points will be at z^2 = - x^2 ( e.g. x =1, z = i).