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I am not professional in Mathematica and I am struggling to understand how the following function acts

$f(x)=\int_{z=0}^\infty \int_{y=0}^{z} \frac{x}{ \sqrt{1-\left(\frac{z^2+y^2-x^2}{2 z y}\right)^2}} e^{- a y^2} \, dy\, e^{-b z^2} \, dz$

where $a>0$ , $b>0$, and obviously $|z-y| \leqslant x \leqslant y+z$.

I tried the following code

Integrate[x*e^(-a*y^2)*e^(-b*z^2))/(sqrt(1-((z^2+y^2-x^2)/(2*z*y))^2)) , {z, 0, Infinity}, {y, 0, z}, Assumptions -> a > 0 && b>0 && \ abs[y-z]\[LessSlantEqual]x\[LessSlantEqual]y+z]

However, I did not get any understandable result! I would appreciate it if anyone can help me to find a closed-form or approximation of this integral. I want to understand how this function varies with respect to $a$ and $b$? Any kind of plot diagram can also help.

Thanks

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    $\begingroup$ You need to fix some basic Mathematica syntax first: 1) abs should be Abs, 2) \[LessSlantEqual] should be <=, 3) e^... should be Exp[...] if you intend e for to be Euler's number, 4) sqrt(...) should be Sqrt[...], 5) you have a stray backslash \ before the abs $\endgroup$
    – flinty
    Jun 17, 2020 at 12:53
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    $\begingroup$ You could do numerical integration with this for a given $x,a,b$: intg[x_?NumericQ, a_?NumericQ, b_?NumericQ] := NIntegrate[ x/Sqrt[1 - ((z^2 + y^2 - x^2)/( 2 z y))^2] Exp[-a y^2] Exp[-b z^2], {z, 0, Infinity}, {y, 0, z}] . I doubt there is a closed form. $\endgroup$
    – flinty
    Jun 17, 2020 at 13:13

1 Answer 1

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Not a real answer, but some graphical overview.

Look, what a and b- factors alone do

integrand[x_, y_, z_, a_, b_] = (x E^(-a y^2) E^(-b z^2))/Sqrt[
      1 - ((z^2 + y^2 - x^2)/(2 z y))^2];

Manipulate[
   Plot3D[E^(-a y^2 - b z^2), {z, 0, 10}, {y, 0, 10}, 
   RegionFunction -> Function[{z, y, int}, 0 <= y <= z], 
   PlotRange -> {0, 1}, PlotPoints -> 50, 
   AxesLabel -> {z, y, int}], {{a, 1}, 0, 4, 
   Appearance -> "Labeled"}, {{b, 1}, 0, 5, 
   Appearance -> "Labeled"}]

Look, what the whole integrand does

Manipulate[
  Plot3D[integrand[x, y, z, a, b], {z, 0, 3}, {y, 0, 3}, 
   RegionFunction -> 
   Function[{z, y, int}, Abs[z - y] <= x <= y + z && 0 <= y <= z], 
 PlotRange -> {0, 4}, PlotPoints -> 50, 
 AxesLabel -> {z, y, int}], {{x, 1}, 0, 5, 
 Appearance -> "Labeled"}, {{a, 1}, 0, 4, 
 Appearance -> "Labeled"}, {{b, 1}, 0, 5, 
 Appearance -> "Labeled"}]
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