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Badger science forum beginner
Joined: 07 May 2006
Posts: 38

Posted: Tue Jul 18, 2006 1:49 pm Post subject:
Re: Integrate erf type expression



On Mon, 17 Jul 2006 18:20:54 EDT, Pratim Vakish
<pratim_usc@hotmail.com> wrote:
Quote:  Thank you very much for your answer Badger.
Just an additional question to be sure I have fully understood.
Let me consider this expression:
I = Int[oo...z] e^1/2*((xy)/0.1)^2 dx
= sqrt(pi)/2 *
Erf(x/(sqrt(2)*0.1)  y/((sqrt(2)*0.1)))  [oo...z]

Not correct. Because of the constants you introduced into the
problem, there will be a new constant factor in the antiderivative.
To see this, let's say you want to evaluate
Int e^( a (xy) )^2 dx
where a is a real constant > 0 (this form is sufficient to do your new
integral).
Make the substitution
t = a (xy)
Then dt = a dx, or dx = dt / a
The integral becomes
Int (1/a) e^t^2 dt = sqrt(pi)/2 (1/a) Erf(t)
Notice the constant factor 1/a, and remember that the sqrt(pi)/2 is
present because of the way Mathematica defines the Erf function.
Now, substitute for t, and we have the antiderivative
sqrt(pi)/2 (1/a) Erf( a (xy) )
For your new problem, you want to evaluate
I = Int[oo...z] e^1/2*((xy)/0.1)^2 dx
which, if I'm reading it correctly, can be written
I = Int[oo...z] e^( (xy) / (0.1 sqrt(2)) )^2 dx
= Int[oo...z] e^( 5 sqrt(2) (xy) )^2 dx
So, using our result above, with a = 5 sqrt(2), we have
I = sqrt(pi)/2 1/(5 sqrt(2)) Erf( 5 sqrt(2) (xy) )  [oo...z]
= sqrt(pi)/2 1/(5 sqrt(2)) [ Erf( 5 sqrt(2) (zy) )  Erf(oo) ]
= sqrt(pi)/2 1/(5 sqrt(2)) [ Erf( 5 sqrt(2) (yz) ) + 1 ]
= sqrt(pi)/2 1/(5 sqrt(2)) [ 1  Erf( 5 sqrt(2) (yz) ) ]
I = sqrt(pi)/2 1/(5 sqrt(2)) Erfc( 5 sqrt(2) (yz) )
Quote: 
= sqrt(pi)/2 [Erf(z/(sqrt(2)*0.1)  y/(sqrt(2)*0.1))
 Erf(ooy/(sqrt(2)*0.1)) ]
Now, Erf is an odd function and Erf(oo) = 1, so
I = sqrt(pi)/2 [ Erf(y / (sqrt(2)*0.1)  z (sqrt(2)*0.1)) + Erf(oo) ]
= sqrt(pi)/2 [ 1  Erf(y/(sqrt(2)*0.1)z/(sqrt(2)*0.1))]
Now if I take the derivative of the expression above with respect to y, is it correct to say that it is equal to:
=  sqrt(pi)/2*2/sqrt(pi) * 1/(sqrt(2)*0.1) *
e^[(y/(sqrt(2)*0.1)z/(sqrt(2)*0.1))]
= 1/(sqrt(2)*0.1) *
e^[(y/(sqrt(2)*0.1)z/(sqrt(2)*0.1))]
Is it correct?
Is ther simplication I can do beforehand?

Not correct. You say that you want the derivative with respect to y?
Why not with respect to z? I'll do it with respect to z.
I' = d/dz sqrt(pi)/2 1/(5 sqrt(2)) Erfc( 5 sqrt(2) (yz) )
First, I have it that
d/dz Erfc( 5 sqrt(2) (yz) )
= 2/sqrt(pi) [ e^( 5 sqrt(2) (yz) )^2 ] (5 sqrt(2))
and so
I' = e^( 5 sqrt(2) (yz) )^2
which is equivalent to the original integrand.
Now, you should do all the work yourself and see if you agree with
these results. 

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Pratim Vakish science forum beginner
Joined: 17 Jul 2006
Posts: 2

Posted: Mon Jul 17, 2006 10:20 pm Post subject:
Re: Integrate erf type expression



Thank you very much for your answer Badger.
Just an additional question to be sure I have fully understood.
Let me consider this expression:
I = Int[oo...z] e^1/2*((xy)/0.1)^2 dx
= sqrt(pi)/2 *
Erf(x/(sqrt(2)*0.1)  y/((sqrt(2)*0.1)))  [oo...z]
= sqrt(pi)/2 [Erf(z/(sqrt(2)*0.1)  y/(sqrt(2)*0.1))
 Erf(ooy/(sqrt(2)*0.1)) ]
Now, Erf is an odd function and Erf(oo) = 1, so
I = sqrt(pi)/2 [ Erf(y / (sqrt(2)*0.1)  z (sqrt(2)*0.1)) + Erf(oo) ]
= sqrt(pi)/2 [ 1  Erf(y/(sqrt(2)*0.1)z/(sqrt(2)*0.1))]
Now if I take the derivative of the expression above with respect to y, is it correct to say that it is equal to:
=  sqrt(pi)/2*2/sqrt(pi) * 1/(sqrt(2)*0.1) *
e^[(y/(sqrt(2)*0.1)z/(sqrt(2)*0.1))]
= 1/(sqrt(2)*0.1) *
e^[(y/(sqrt(2)*0.1)z/(sqrt(2)*0.1))]
Is it correct?
Is ther simplication I can do beforehand? 

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Badger science forum beginner
Joined: 07 May 2006
Posts: 38

Posted: Mon Jul 17, 2006 4:16 pm Post subject:
Re: Integrate erf type expression



On Mon, 17 Jul 2006 10:27:35 EDT, Pratim Vakish
<pratim_usc@hotmail.com> wrote:
Quote:  My question is probably very basic. I am a novice in the field.
I would like to compute/simplify the following expression:
Integrale [from infinity to z] of [e^(xy)^2]dx
Could you please help?
Best

In what follows, I have used Mathematica's definitions of Erf and Erfc
as given here:
<http://mathworld.wolfram.com/Erf.html>
<http://mathworld.wolfram.com/Erfc.html>
Some authors define these functions without the factor 2/sqrt(pi).
Given
I = Int[oo...z] e^(xy)^2 dx
= sqrt(pi)/2 Erf(xy)  [oo...z]
= sqrt(pi)/2 [ Erf(zy)  Erf(ooy) ]
Now, Erf is an odd function and Erf(oo) = 1, so
I = sqrt(pi)/2 [ Erf(yz) + Erf(oo) ]
= sqrt(pi)/2 [ 1  Erf(yz) ]
I = sqrt(pi)/2 Erfc(yz)
The 'erf' functions can be evaluated numerically using Mathematica
here:
<http://functions.wolfram.com/GammaBetaErf/>
Click on Erfc, then 'Evaluation'.
HTH 

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Peter Spellucci science forum Guru
Joined: 29 Apr 2005
Posts: 702

Posted: Mon Jul 17, 2006 3:37 pm Post subject:
Re: Integrate erf type expression



In article <13642928.1153146486183.JavaMail.jakarta@nitrogen.mathforum.org>,
Pratim Vakish <pratim_usc@hotmail.com> writes:
Quote:  My question is probably very basic. I am a novice in the field.
I would like to compute/simplify the following expression:
Integrale [from infinity to z] of [e^(xy)^2]dx
Could you please help?
Best

what about a little substitution to get it into
const* integral_{infinity to somewhere} exp(t^2/2) dt
then using the normal distribution function?
hth
peter 

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Pratim Vakish science forum beginner
Joined: 17 Jul 2006
Posts: 2

Posted: Mon Jul 17, 2006 2:27 pm Post subject:
Integrate erf type expression



My question is probably very basic. I am a novice in the field.
I would like to compute/simplify the following expression:
Integrale [from infinity to z] of [e^(xy)^2]dx
Could you please help?
Best 

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