My problem is that I am trying to find a Numerical solution to a function of the form f(x,c)== K for x for some given values of c, K, and plot this function over a wide range of values of c.

My function f(x,c) is of the form Integrate[g(x,y,c),{y,a,b}], where a and b are some numerical values. This integration however cannot be evaluated symbolically in x and c,as g is a really complicated function and hence can only be performed numerically.

So if I write my function as f(x_,c_)= NIntegrate[g(x,y,c),{y,a,b}], I get some numerical value of f(x,c) for some numerical values of x and c. (If I write it as Integrate[g(x,y,c),{y,a,b}] I donot get any value for f(x,c)).

But with this form of the function, I cannot get any solution using NSolve, or FindRoot. (FindRoot doesn't converge, and it may not be practical if I plan to plot it for a wide range of values of c).

So at this point, plotting the solutions seems like a far fetched dream. I apologize for not providing details of the functions, as the function g arises from a lot of complicated functions.

Thanks in advance for your help


I decided to post my attempt, and some extent of the functions, based on one of the Solutions by Carl. For my case arh is a known function, and gamma and w are numerical values:

eqn = Inactive[Integrate][
 g1[Hi[mx], a, mx]*mx, {a, 1, arh[\[Gamma], w]}] + 
 g2[Hi[mx], a, mx]*mx, {a, arh[\[Gamma], w], \[Infinity]}] ==0.1198/(9.2*10^24);

Distribute /@ D[eqn, mx]

int0[Hi_, mx_?NumberQ] :=NIntegrate[g1[Hi, a, mx], {a, 1, arh[\[Gamma], w]}]
int1[Hi_, mx_?NumberQ] := 
  Derivative[0, 0, 1][g1][Hi, a, mx], {a, 1, arh[\[Gamma], w]}]
int2[Hi_, mx_?NumberQ] := 
  Derivative[1, 0, 0][g1][Hi, a, mx], {a, 1, arh[\[Gamma], w]}]

int02[Hi_, mx_?NumberQ] := 
 NIntegrate[g2[Hi, a, mx], {a, arh[\[Gamma], w], \[Infinity]}]
int12[Hi_, mx_?NumberQ] := 
  Derivative[0, 0, 1][g2][Hi, a, mx], {a, 
   arh[\[Gamma], w], \[Infinity]}]
int22[Hi_, mx_?NumberQ] := 
  Derivative[1, 0, 0][g2][Hi, a, mx], {a, 
   arh[\[Gamma], w], \[Infinity]}]

My Functions are of the form

g1[Hi_, a_, mx_] := 
  a^2/(Hrh[a, \[Gamma], w, Hi]*Trh[\[Gamma], Hi]^3) \[Sigma]vtotal[
    a, \[Gamma], w, Hi, mx, Trh11] nXeq[a, \[Gamma], w, Hi, mx, 
g2[Hi_, a_, mx_] := 
  a^2/(H[a, \[Gamma], w, Hi]*Trh[\[Gamma], Hi]^3) \[Sigma]vtotal[
    a, \[Gamma], w, Hi, mx, T] nXeq[a, \[Gamma], w, Hi, mx, T]^2;

And I try to Solve:

sol = NDSolveValue[{(int0[Hi[mx], a, mx] + 
        mx (int1[Hi[mx], mx] + 
           Derivative[1][Hi][mx] int2[Hi[mx], mx])) + (int02[Hi[mx], 
         a, mx] + 
        mx (int12[Hi[mx], mx] + 
           Derivative[1][Hi][mx] int22[Hi[mx], mx])) == 0, 
    Hi[10^-7] == 10^-9}, Hi, {mx, .1, 10^-19}];

This returns an error NDSolveValue::ndnum: Encountered non-numerical value for a derivative at mx == 1.`*^-7.

NOTE: Trh, and \sigmavtotal and nXeq are defined previously, and are of the form: \[Sigma]vtotal[a_, \[Gamma]_, w_, Hi_, mx_, T_Symbol] nXeq[a_, \[Gamma]_, w_, Hi_, mx_, T_Symbol] Trh[\[Gamma]_, Hi_]

and the other functions are of the form T[a_, \[Gamma]_, w_, Hi_] and Trh11[a_, \[Gamma]_, w_, Hi_]

I want to plot for mx: 0.1 to 10^-19

  • 1
    $\begingroup$ Share your attempt otherwise we will be not of much help. $\endgroup$
    – zhk
    Oct 13, 2018 at 4:12
  • $\begingroup$ Recommend you provide code that fully runs to where we can replicate your error (I get a different error due to arh not a valid limit of integration) $\endgroup$
    – MikeY
    Dec 13, 2018 at 1:06

2 Answers 2


Without a concrete example I can only guess at an approach that might work.

g[x_, y_, c_] := 1/Sqrt[x^2 + y^2 + c^2]

a = 0; b = 1;

Functions that use numeric techniques should have their arguments restricted to numeric values.

f[x_?NumericQ, c_?NumericQ] := NIntegrate[g[x, y, c], {y, a, b}]

soln[c_?NumericQ, K_?NumericQ] := x /. FindRoot[f[x, c] == K, {x, 1/2}]

Verifying that soln produces a numeric value

soln[1/2, 1/2]

(* 1.85275 *)

Plotting soln for a range of c and K values

Plot3D[soln[c, K], {c, 0, 1}, {K, 1/10, 1}]

enter image description here

  • $\begingroup$ I posted more details of my specific question. Thanks in Advance $\endgroup$
    – TheInvoker
    Oct 13, 2018 at 19:24

I would use NDSolveValue for this kind of problem. You are trying to find the dependence of x on c, so let's use x[c] to make this dependence explicit. Then your equation is:

eqn = Inactive[Integrate][g[x[c], y, c], {y, 1, 2}] == k;

where I arbitrarily chose $a=1$ and $b=2$. Let's differentiate this equation to obtain an ODE:

Distribute /@ D[eqn, c] //TeXForm

$\int _1^2x'(c) g^{(1,0,0)}(x(c),y,c)dy+\int _1^2g^{(0,0,1)}(x(c),y,c)dy=0$


int0[x_, c_?NumberQ] := NIntegrate[g[x, y, c], {y, 1, 2}]
int1[x_, c_?NumberQ] := NIntegrate[Derivative[0,0,1][g][x, y, c], {y, 1, 2}]
int2[x_, c_?NumberQ] := NIntegrate[Derivative[1,0,0][g][x, y, c], {y, 1, 2}]

where the _?NumberQ test prevents premature evaluations of NIntegrate with non-numeric integrands. Finally, let's come up with a sample g:

g[x_, y_, c_] := BesselI[1, x y] BesselJ[1, c y]

Then, the ODE to be solved is:

sol = NDSolveValue[
    int2[x[c], c] x'[c] + int1[x[c], c] == 0,
    x[1] == 1
    {c, .1, 2}

Let's check whether sol satisfies the required equation:

Plot[int0[sol[c], c], {c, .1, 2}, PlotRange -> 1]

enter image description here

And a visualization of the solution:

Plot[sol[c], {c, .1, 2}]

enter image description here

  • $\begingroup$ This solution looks really promising. The only problem is that I donot know of any initial condition like x[1]==1 . I can guess an order, like x[10^-7]==10^-10. Here is what my eqn looks like:= eqn = Inactive[Integrate][ g1[Hi[mx], a, mx]*mx, {a, 1, arh[[Gamma], w]}] + Inactive[Integrate][ g2[Hi[mx], a, mx]*mx, {a, arh[[Gamma], w], [Infinity]}] == 0.1198/(9.2*10^24); There are two seperate complicated functions g1 and g2 over different known ranges, gamma and w are known, and arh is a known function. I need to find Hi[mx], and I need to plot from mx= 0.1 to 10^-19 $\endgroup$
    – TheInvoker
    Oct 13, 2018 at 18:16
  • $\begingroup$ I posted more details by editing the question, thanks in advance $\endgroup$
    – TheInvoker
    Oct 13, 2018 at 19:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.