Undergrad Is my integral integrable using Mathematica or is there a fundamental error?

[Hawkeye18]

This might simplify your calculations a bit: integrating by parts you get $$\int_0^\infty e^{-x} f_X(x) dx = \int_0^\infty F_X(x) e^{-x} dx, $$ so you do not need to compute derivative of $F_X$.

 

[EngWiPy]

Oh, really? So, I think you can see now where the integral in my first post came from. I will follow on this.

 

[EngWiPy]

Integration by parts goes like this. Let $u=e^{-x}$ and thus $du=-e^{-x}\,dx$, and $dv=f_X(x)\,dx$, and thus $v=\int f_X(x)\,dx$. I know that $F_X(x)=\int_0^xf_X(x)\,dx$, but does the above $v=F_X(x)$ and why? I will continue. Assuming the above is correct, we have

\int_0^{\infty}e^{-x}f_X(x)\,dx=\underbrace{\left. e^{-x}F_X(x)\right|_0^{\infty}}_{=0}+\int_0^{\infty}e^{-x}F_X(x)\,dx=\int_0^{\infty}e^{-x}F_X(x)\,dx.

Interesting!

 

[EngWiPy]

Based on the above derivations, the average value of $\varepsilon(\alpha_1,\alpha_2,\alpha_3)$ is given by

\varepsilon=1-\frac{A}{B}\int_0^{\infty}x^{-1}\exp\left[\frac{x+A}{xB}-x\right]E_1\left[\frac{x+A}{xB}\right]\,dx

Is the above integral inetgrable now?

 

[EngWiPy]

OK, now the integral is integrable, and Mathematica gives no complains. But when I compared the numerical results with Monte-Carlo simulations, where I generated $10^5$ samples for each point, I got very close but exactly the same curves. See attached Figure. My original equation is $\varepsilon(\alpha_1,\alpha_2\alpha_3)=0.5\exp\left(-\frac{\frac{\alpha_1}{\alpha_2}G\gamma_Q}{\frac{1}{G}\alpha_3\gamma_p+1}\right)$. So,

\varepsilon=0.5-0.5\frac{G^2\gamma_Q}{\gamma_p}\int_0^{\infty}x^{-1}\exp\left[\frac{G\left(x+G\gamma_Q\right)}{x\gamma_p}-x\right]E_1\left[(\frac{G\left(x+G\gamma_Q\right)}{x\gamma_p}\right]\,dx

The Mathematica code for the above equation is

yp = 10^(0/10);
GSS = 50;
For[yQdB = -10, yQdB <= 15, yQdB++;
yQ = 10^(yQdB/10);
A1 = 0.5 -
   0.5*((GSS^2)*yQ )/yp*
    NIntegrate[
     1/x*Exp[(GSS*(x + GSS*yQ))/(x*yp) - x]*
      ExpIntegralE[1, (GSS*(x + GSS*yQ))/(x*yp)], {x, 0, Infinity},
     PrecisionGoal -> 5, MaxRecursion -> 20];
Print[A1];]

I did Monte-Carlo simulations as follows

1. For each $\gamma_Q$ generate three exponential random variables $\alpha_i$ for i=1,2,3.
2. Find the value of $\varepsilon(\alpha_1,\alpha_2\alpha_3,\gamma_Q)=\varepsilon(\alpha_1,\alpha_2\alpha_3,\gamma_Q)+0.5\exp\left(-\frac{\frac{\alpha_1}{\alpha_2}G\gamma_Q}{\frac{1}{G}\alpha_3\gamma_p+1}\right)$, where the initial value of $\varepsilon(\alpha_1,\alpha_2\alpha_3,\gamma_Q)$ is zero.
3. Repeat for $N=10^5$ iterations.
4. Find the average value as $\varepsilon(\gamma_Q)=\varepsilon(\alpha_1,\alpha_2\alpha_3,\gamma_Q)/N$

Is there something wrong that gives me the difference between the two curves?

I didn't know how to attach the figure!

 

[EngWiPy]

Is it an error margin in the numerical evaluation of the integral?

 

[Hawkeye18]

Can you show the figures? What error do you have?

 

[EngWiPy]

How to upload a figure from my PC?

 

[Hawkeye18]

Upload image to a hosting site (dropbox, google photos, tumblr, flickr, etc) then click to "image" on the toolbar and insert the image url.

 

[EngWiPy]

It doesn't work. I tried both dropbox and google photos. Isn't it "get a link" that I need to insert here?

 

[EngWiPy]

Attached is the figure. Sim=Monte-Carlo simulations

 

[EngWiPy]

I think Monte-Carlo simulations are accurate, so, I suspect, given that everything else done properly, the accuracy of NIntegrate in Mathematica is the issue. Is there any possible reason and I cannot see it?

 

[Hawkeye18]

Monte-Carlo convergence is quite slow, $C/\sqrt N$, maybe the error is due to that. Also there could be some details that Monte-Carlo misses.
On the other hand how Mathematica does NIntegrate is hidden, it might also introduce some systematic error here.

 

[EngWiPy]

I remember using Mathematica to evaluate some numerical integral in my master thesis. There was no difference between the numerical integration and Monte-Carlo simulations then!