Help me compute numerically this integral

[bayes]

Hi,

Can anybody with a powerful computer help me compute numerically the following integral:

1) k[x_, y_] :=
Exp[-Integrate[(Cosh[2 t] Cos[
2 t (ArcSinh[Cot[x]] - ArcSinh[Cot[y]])/Pi] - 1) Exp[
t]/(t Sinh[2 t] Cosh[t]), {t, 0, Infinity}]]

2) NIntegrate[k[x, y], {x, -PI, 0}, {y, -PI, 0}]

I'm using the notation of mathematica. I am interested in in the integral defined in 2)

This is extremely important to me and any help is highly appreciated.

Thanks

 

[jackmell]

What happens when you plug-in the NIntegrate command into Wolfram Alpha? Do a google search on it if you're not familiar. It's a internet interface to (some of) the features of Mathematica.

 

[bayes]

Thank you for your reply but Mathematica alpha does only single integrals whereas mine involves 3 integrals. Moreover mathematica isn't capable of computing this integral.

 

[berkeman]

Hi,

Can anybody with a powerful computer help me compute numerically the following integral:

1) k[x_, y_] :=
Exp[-Integrate[(Cosh[2 t] Cos[
2 t (ArcSinh[Cot[x]] - ArcSinh[Cot[y]])/Pi] - 1) Exp[
t]/(t Sinh[2 t] Cosh[t]), {t, 0, Infinity}]]

2) NIntegrate[k[x, y], {x, -PI, 0}, {y, -PI, 0}]

I'm using the notation of mathematica. I am interested in in the integral defined in 2)

This is extremely important to me and any help is highly appreciated.

Thanks

What is the context of this integral? Where does it come from? What is the application? Is this a school project?

 

[bayes]

This integral shows up in a research project in physics. Unfortunately I can't give you more details than that.

Thanks

 

[jackmell]

I think you need to more analyze the integrand in the domain of integration. For example, if I just choose a point (x,y)=(-pi/4,-pi/4) and then attempt to numerically integrate over t this is what Mathematica reports:

In[1]:=
myk[x_, y_, t_] := 
  ((Cosh[2*t]*Cos[(2*t*(ArcSinh[Cot[x]] - ArcSinh[Cot[y]]))/Pi] - 1)*
    Exp[t])/(t*Sinh[2*t]*Cosh[t])

In[3]:=
myx = -Pi/4
myy = -Pi/4
NIntegrate[myk[myx, myy, t], {t, 0, Infinity}]

Out[3]=
-(Pi/4)

Out[4]=
-(Pi/4)

During evaluation of In[3]:= NIntegrate::inumri:
The integrand (E^t (-1+Cosh[2 t]) Csch[2 t] Sech[t])/t has 
evaluated to Overflow, Indeterminate, or Infinity for all 
sampling points in the region with 
boundaries {{0.,4.64782*10^14}}. >>

Out[5]=
NIntegrate[myk[myx, myy, t], {t, 0, Infinity}]

 

[bayes]

Thanks jackmell for your help but I think mathematica is not able to evaluate this integral. I know for a fact that when x = y then the integral over t is zero so I don't understand your result when x=y = -PI/4. If anyone knows another numerical method such as Gauss quadrature or Monte Carlo that would be great.

 

[jackmell]

Wait a minute dude. Let's just plot it then:

Plot[myk[-Pi/4, -Pi/4, t], {t, 0, 100}, PlotRange -> {{0, 100}, {0, 2}}]

and also, use FindInstance to see if it ever goes below the x-axis:

In[23]:= FindInstance[
 myk[-\[Pi]/4, -\[Pi]/4, t] == 0 && t > 0, t, Reals]

Out[23]= {}

so at least initially, these results are suggesting the integrand when (x,y)=(-pi/4,-pi/4) is always positive and note if I just start integrating it from t=0, 100, then 1000, then 10000, then 100000, the value of the integral grows larger and larger. These results are leading me to believe the integral that I have input to Mathematica may indeed not converge but there is still a slight chance that Mathematica is wrong but it's not often.

 

[bayes]

Yes I think when x = y your integral goes to +Infinity and therefore Exp[-Integral] goes to 0 and that's what I mean when I said I expect it to go to zero when x = y. I really meant the exponential of minus the integral. So our calculations agree when x = y. Thanks for your help.

 

[jackmell]

Yes I think when x = y your integral goes to +Infinity and therefore Exp[-Integral] goes to 0 and that's what I mean when I said I expect it to go to zero when x = y. I really meant the exponential of minus the integral. So our calculations agree when x = y. Thanks for your help.

Ok, sorry. Forgot about the Exp. Yeah, looks like a challenging integral but I have confidence the Mathematica experts at Drexel could do it. So if you really need more help, then send them an e-mail about it:

http://mathforum.org/kb/forum.jspa?forumID=79