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How to compute this integral ?

by y33t
Tags: compute, integral
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y33t
#1
Dec4-12, 03:25 PM
P: 107
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.
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arildno
#2
Dec4-12, 03:29 PM
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Letting "x" be the variable to integrate first, set x=z*Sinh(u), where Sinh(u) is the hyperbolic sine function.
y33t
#3
Dec4-12, 03:57 PM
P: 107
Quote Quote by arildno View Post
Letting "x" be the variable to integrate first, set x=z*Sinh(u), where Sinh(u) is the hyperbolic sine function.
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
#4
Dec4-12, 04:05 PM
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How to compute this integral ?

Quote Quote by y33t View Post
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 ?
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
#5
Dec4-12, 04:08 PM
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x=z tan t can also do the trick.
arildno
#6
Dec4-12, 04:11 PM
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Quote Quote by dextercioby View Post
x=z tan t can also do the trick.
Sure enough.
Scribbling out the solution with tan(t) or Sinh(t) takes about the same amount of time and space, though..
arildno
#7
Dec5-12, 09:03 AM
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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
#8
Dec6-12, 09:25 AM
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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
#9
Dec6-12, 02:13 PM
P: 9
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).


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