Author 
Message 
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 

Back to top 


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 

Back to top 


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 

Back to top 


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? 

Back to top 


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. 

Back to top 


Google


Back to top 



The time now is Thu Apr 26, 2018 11:25 am  All times are GMT

