# How to compute this integral ?

1. Dec 4, 2012

### 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

2. Dec 4, 2012

### arildno

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

3. Dec 4, 2012

### 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 ?

4. Dec 4, 2012

### 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.

5. Dec 4, 2012

### dextercioby

x=z tan t can also do the trick.

6. Dec 4, 2012

### arildno

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

7. Dec 5, 2012

### arildno

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

8. Dec 6, 2012

### 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 $sin^2(\theta)+ cos^2(\theta)= 1$ and, dividing through by $cos^2(\theta)$, $tan^2(\theta)+ 1= sec^2(\theta)$.

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

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

9. Dec 6, 2012

### 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).

Last edited: Dec 6, 2012