I want to obtain $I$ as a function of $V$, in the following equation

$I = \int\limits_0^\infty\int\limits_0^\infty \left(\frac{1}{1 + e^{x+y}}\right) \left(\frac{1}{e^{V-x-y}+1}\right) \left(\frac{1}{\sqrt{y}}\right) dy\ dx$

I want to numerically evaluate the integral so as to get a curve of $I$ versus $V$.

I used the following commands

f[v_] := 
    (1/(1 + Exp[x + y])) (1/(Exp[v - x - y] + 1)) (1/Sqrt[y]), 
    {y, 0, 1000000}, {x, 0, 1000000}]

Plot[f[v], {v, 0, 1}]

It gives me the error message:

NIntegrate::izero: Integral and error estimates are 0 on all integration subregions. Try increasing the value of the MinRecursion option. If value of integral may be 0, specify a finite value for the AccuracyGoal option.

Context - Basically, the integrand is a product of fermi-functions which i am trying to evaluate so as to get current versus voltage relation in a device.


closed as off-topic by Anton Antonov, Henrik Schumacher, José Antonio Díaz Navas, MarcoB, Sektor Apr 12 '18 at 18:20

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  • 2
    $\begingroup$ 1. Use SetDelayed[] (:=) instead of Set[] (=) when defining your function. 2. Reduce your function to a single NIntegrate[]. Thus: f[v_] := NIntegrate[(1/(1 + Exp[x + y])) (1/(1 + Exp[v - x - y])) (1/Sqrt[y]), {x, 0, ∞}, {y, 0, ∞}] Alternatively: f[v_] := NIntegrate[LogisticSigmoid[x + y - v] LogisticSigmoid[-x - y]/Sqrt[y], {x, 0, ∞}, {y, 0, ∞}] $\endgroup$ – J. M. will be back soon Mar 7 '18 at 21:34
  • $\begingroup$ For comparison, the exact integral can be evaluated for v == 0, i.e, Integrate[ (1/(1 + Exp[x + y])) (1/(Exp[-x - y] + 1)) (1/Sqrt[y]), {y, 0, Infinity}, {x, 0, Infinity}] evaluates to (-(-1 + Sqrt[2]))*Sqrt[Pi]*Zeta[1/2] or 1.072154929940191 $\endgroup$ – Bob Hanlon Mar 8 '18 at 0:42

One can do the x integral analytically:

g[v_] = Integrate[
    (1/(1+Exp[x+y])) (1/(Exp[v-x-y]+1)) (1/Sqrt[y]),
    {x, 0, Infinity},
    Assumptions -> y>0 && v>0

Log[(E^v + E^y)/(1 + E^y)]/((-1 + E^v) Sqrt[y])

Using the above we can define:

f[v_] := NIntegrate[g[v], {y, 0, Infinity}]


Plot[f[v], {v, 0, 1}]

enter image description here

  • $\begingroup$ The answer given by a single NIntegrate and your method have orders of magnitude difference. $\endgroup$ – Indeterminate Mar 7 '18 at 23:35
  • $\begingroup$ @JashanSinghal You should give an example v that demonstrates your claim. Perhaps the issue is that you chose to use 1000000 as an upper limit instead of Infinity. $\endgroup$ – Carl Woll Mar 7 '18 at 23:52

Direct numerical solution is

f[v_?NumericQ] :=NIntegrate[(1/(1 + Exp[x + y])) (1/(Exp[v - x - y] + 1)) (1/Sqrt[y]), {y, 0, Infinity}, {x, 0, Infinity}]

Plot[f[v], {v, 0, 1}]

enter image description here


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