# Eliminate a variable within an integral

I have two expressions given by,

$$a = \int_0^1 dx \frac{c z_s^{d+1} x^d}{\sqrt{(1-(z_s/z_h)^{d+1} x^{d+1})(1-c^2 z_s^{2d} x^{2d})}} \tag{1}\label{1},$$

$$S = \int_\epsilon^{z_s} dx \frac{1}{x^d} \sqrt{\frac{1}{(1-(x/z_h)^{d+1})(1-c^2 x^{2d})}}\tag{2}\label{2},$$

where $$c=c(z_s)$$ and $$a$$ is a function of $$z_s$$, and $$z_h$$ is a constant which can be given a value later. Also, even though $$a$$ is a function of $$z_s$$ in the expression, in the final expression for $$S$$ I will just give it a value. $$d$$ is just a dimension parameter which can be given a value say, $$d=3$$.

The goal is to find an expression of $$S$$ such that $$c$$ is eliminated. Now, one condition which can help achieve that is,

$$\frac{dS}{dz_s} = 0\tag{3}\label{3},$$

actually I need to find an expression for the minimal $$S$$ (with respect to $$z_s$$) to which $$c$$ dictates.

I have asked some related problems before but for eq. $$\eqref{2}$$ in those cases I made a transformation $$x=z_sy$$ so that the integral limits there is $$[\epsilon/z_s,1]$$ and then expand out the finite and divergent terms then I just showed the finite terms there so that I allow the lower limit to be 0.

Solving a function inside an integral

Smoothen the result of FindRoot

In this case, I just start with the original integral $$\eqref{2}$$ and set a small number for $$\epsilon$$ (this is just a modification of my previous problem in order to eliminate the divergence contribution but the result should not differ for those regions far from the divergence region).

The problem is when I try to plot those same quantities I get a different result.

The previous code,

d = 3;
zh = 1.5;
torootL[a_, c_?NumericQ, z_] := a - NIntegrate[(c z^(d + 1) xd^(d/(d + 1)))/((1 - xd (z/zh)^(d + 1)) (1 - xd^((2 d)/(d + 1)) c^2 (z)^(2 d)))^(1/2), {xd, 0, 1}, MaxRecursion -> 12, PrecisionGoal -> 6, Method -> {"GlobalAdaptive", "SingularityHandler" -> "DoubleExponential"}]
cz[a_?NumericQ, z_?NumericQ] := c /. FindRoot[torootL[a, c, z], {c, 0.001, 0.0000001, 1000}]
intS[a_?NumericQ, z_?NumericQ] := NIntegrate[With[{b = z/zh}, (((-1)/(d - 1)) cz[a, z]^2 z^(2 d)) x^d ((1 - (b x)^(d + 1))/(1 - cz[a, z]^2 (z x)^(2 d)))^(1/2) - ((b^(d + 1) (d + 1))/(2 (d - 1))) x ((1 - cz[a, z]^2 (z x)^(2 d))/(1 - (b x)^(d + 1)))^(1/2) + (b^(d + 1) x)/((1 - (b x)^(d + 1)) (1 - cz[a, z]^2 (z x)^(2 d)))^(1/2)], {x, 0, 1}, MaxRecursion -> 12, PrecisionGoal -> 6, Method -> {"GlobalAdaptive", "SingularityHandler" -> "DoubleExponential"}]
functionS[a_, z_] = ((-((1 - cz[a, z]^2 z^(2 d)) (1 - (z/zh)^(d + 1)))^(1/2)/(d - 1)) + intS[a, z] + 1)/(4 z^(d - 1));
Plot[functionS[0.1, z], {z, 1, zh}, PlotPoints -> 8, AxesLabel -> {"z_s", "S"}, AxesStyle -> Directive[Black, 18], PlotStyle -> Thick, PlotRange -> Full, ImageSize -> Large] // Quiet // AbsoluteTiming


produces this plot, The present code where I just changed the lines corresponding to the discussion I made above,

d = 3;
zh = 1.5;
torootL[a_, c_?NumericQ, z_] := a - NIntegrate[(c z^(d + 1) xd^(d/(d + 1)))/((1 - xd (z/zh)^(d + 1)) (1 - xd^((2 d)/(d + 1)) c^2 (z)^(2 d)))^(1/2), {xd, 0, 1}, MaxRecursion -> 12, PrecisionGoal -> 6, Method -> {"GlobalAdaptive", "SingularityHandler" -> "DoubleExponential"}]
cz[a_?NumericQ, z_?NumericQ] := c /. FindRoot[torootL[a, c, z], {c, 0.001, 0.0000001, 1000}]

intS[a_?NumericQ, z_?NumericQ] := NIntegrate[(1/x^d) (1/((1 - (x/zh)^(d + 1)) (1 - cz[a, z]^2 x^(2 d))))^(1/2), {x, 10^-10, z}, MaxRecursion -> 20, PrecisionGoal -> 6, Method -> {"GlobalAdaptive"}]
functionS[a_, z_] = intS[a, z] + 1/(z^(d - 1));
Plot[functionS[0.1, z], {z, 1, zh}, PlotPoints -> 8, AxesLabel -> {"z_s", "S"}, AxesStyle -> Directive[Black, 18], PlotStyle -> Thick, PlotRange -> Automatic, ImageSize -> Large] // Quiet // AbsoluteTiming


produces the plot, where the Method I used in the present is only GlobalAdaptive since if I use the Method in the old code it produces a flat line which I don't know why. Anyways, the old and present codes should produce the same plot especially in those region close to $$z_s = 1.5$$ to which in the first image you can see a minimum occurring around $$z_s \approx 1.4$$.

Why is this so? I'm still trying to understand the NIntegrate Integration Strategies documentation although it is hard to understand for me since I'm not a heavy user maybe until now.

I was thinking of maybe there is a different strategy to eliminate $$c$$ in the expression for $$S$$ using $$\eqref{1}$$ $$\eqref{2}$$ $$\eqref{3}$$ at the same time. Any comments?

UPDATE:

I noticed that when I changed the lower limit of the integral in intS I can get the same curve but with different magnitude so I think that cannot be correct.

intS[a_?NumericQ, z_?NumericQ] := NIntegrate[(1/x^d) (1/((1 - (x/zh)^(d + 1)) (1 - cz[a, z]^2 x^(2 d))))^(1/2), {x, 10^-2, z}, MaxRecursion -> 20, PrecisionGoal -> 6, Method -> {"GlobalAdaptive"}] intS[a_?NumericQ, z_?NumericQ] := NIntegrate[(1/x^d) (1/((1 - (x/zh)^(d + 1)) (1 - cz[a, z]^2 x^(2 d))))^(1/2), {x, 1, z}, MaxRecursion -> 20, PrecisionGoal -> 6, Method -> {"GlobalAdaptive"}] You can see that when I changed the lower limit to 10^-2 the curve already appeared but the magnitude is wrong, and when I set it to 1 the curve is there and the magnitude is closer to the original. However, I think I cannot just tune the lower limit, it must be the case that $$\epsilon \leq z_s \leq z_h$$ since $$\epsilon$$ is a regulator. Note that since that is the case, $$z_s$$ should approach $$\epsilon$$ first then take the limit where $$\epsilon \rightarrow 0$$. So in the case of Mathematica I can just set $$\epsilon$$ to the machine-wise zero, I think $$10^{-10}$$ is considered to be 0 but not really setting 0.

• It is true that $\frac{d a}{d z_s} = 0$ ? – Cesareo Oct 13 '20 at 9:52
• @Cesareo Good question, I'm thinking that $a$ should depend on $z_s$ as seen from $\eqref{1}$, although in the end I can just give it a value. I think I should change it to $a=a(z_s)$. – mathemania Oct 13 '20 at 9:58
• It looks like this problem has no unique solution. May be we need to add some restrictions? – Alex Trounev Oct 16 '20 at 11:49
• @AlexTrounev The only restrictions in the problem are that $S$ must be a minimum for some $z_s$ and $c$ is constrained by this fact. – mathemania Oct 18 '20 at 14:42
• @mathemania Are $a, c, z_s$ real? If $a, c, z_s$ are real then we have some restrictions for $c, z_s$ due to convergence of integral. – Alex Trounev Oct 18 '20 at 14:54

If all parameters $$a,z_s,z_h,c$$ are real, then we have constraints dictated by existence of integral $$a(z_s,z_h,c)$$. As an example let put $$d=3, z_h=3/2, a=1/10$$, from this data we can compute $$c(z_s)$$ as follows (we made substitution $$y=x^{d+1})$$:

d = 3; zh = 3/2;
a[zs_?NumericQ, c_?NumericQ] :=
c zs^(d + 1)/(d + 1) NIntegrate[
Sqrt[1/(1 - (zs/zh)^(d + 1) y)/(1 -
c^2 zs^(2 d) y^(2 d/(d + 1)))], {y, 0, 1}]


Since a[] is real, we can plot this function with using constrains for existence of integral

ContourPlot[(a[zs, c] - 1/10) Boole[c^2 zs^(2 d) <= 1] == 0, {zs, 0,
zh}, {c, 0, 1}, PlotPoints -> 50] The line on this picture is function $$c(z_s)$$ we got from the existence of integral. Now we can calculate this function directly. Let check that

 cc[zs_] :=
c /. FindRoot[(a[zs, c] - 1/10) Boole[c^2 zs^(2 d) <= 1] == 0, {c, 0,
1}, Method -> "Brent"]

cc[.781598]

Out[]= 0.99999


This is starting point we use in a loop

zs = .79;
cr = cc[zs]; Do[zs = zs + .01;
cr[i] = c /.
FindRoot[(a[zs, c] - 1/10) Boole[c^2 zs^(2 d) <= 1] == 0, {c,
cr[i - 1]}];, {i, 1, (zh - 79/100) 100}]

lst1 = Table[{(79 + i) .01, cr[i]}, {i, 0, 150 - 79}];


Plot this list and compare with Figure1 above

ListPlot[Join[{{.781598, 1}}, lst1],
AxesLabel -> {"\!$$\*SubscriptBox[\(z$$, $$s$$]\)", "c"}] Now we can interpolate data and use it to compute integral $$S(z_s)$$

fc = Interpolation[Join[{{.781598, 1}}, lst1]];

S[zs_?NumericQ] :=
NIntegrate[
x^-d/Sqrt[(1 - (x/zh)^(d + 1)) (1 - fc[zs]^2 x^(2 d))], {x, .782,
zs}]


Finally we plot $$S(z_s)$$ and check that there is no local minimum with $$S'(z_s)=0$$. Also pay attention that due to $$c(z_s)$$ definition we define $$\epsilon$$ as $$\epsilon =0.782$$

Plot[S[zs], {zs, 0.782, zh},
AxesLabel -> {"\!$$\*SubscriptBox[\(z$$, $$s$$]\)", "S"}] Now we can extend this research for general case when a follows a list, for example

d = 3; zh = 3/2; a0 = {1/20, 1/10, 1/5};
a[zs_?NumericQ, c_?NumericQ] :=
c zs^(d + 1)/(d + 1) NIntegrate[
Sqrt[1/(1 - (zs/zh)^(d + 1) y)/(1 -
c^2 zs^(2 d) y^(2 d/(d + 1)))], {y, 0, 1}];
Table[plot[i] =
ContourPlot[(a[zs, c] - a0[[i]]) Boole[c^2 zs^(2 d) <= 1] == 0, {zs,
0, zh}, {c, 0, 1}, PlotPoints -> 50,
FrameLabel -> {"\!$$\*SubscriptBox[\(r$$, $$s$$]\)", "c"},
PlotLabel -> a0[[i]]], {i, 3}] We can extract data directly from plot[i] as follows

point = Table[
plot[i] //
Cases[#, GraphicsComplex[points_, ___] :> points, Infinity] &, {i,
Length[a0]}];


Using point we can define c=f[i][rs] and S[rs,i], and plot it as

Table[f[i] = Interpolation[point[[i]] // First], {i, Length[a0]}];

x0[i_] := Last[point[[i]] // First] // First

S[zs_?NumericQ, i_] :=
NIntegrate[
x^-d/Sqrt[(1 - (x/zh)^(d + 1)) (1 - f[i][zs]^2 x^(2 d))], {x, x0[i],
zs}]

Table[Plot[S[zs, i], {zs, x0[i], zh},
AxesLabel -> {"\!$$\*SubscriptBox[\(z$$, $$s$$]\)", "S"},
PlotLabel -> a0[[i]]], {i, Length[a0]}] Note, this solution is available due to constraints follow from the suggestion that a, rh,rs,c are real only.