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

$$I(V) = \int _0 ^\infty\int _0 ^\infty \frac{1}{(1 + e^{x+y})(e^{V-x-y}+1)}\frac{1}{\sqrt{y}}dx dy$$

I would prefer to do this on Mathematica if it is easy. I want to numerically evaluate the integral so as to get a curve of I v/s V.

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.

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    $\begingroup$ There are three $($s and just two $)$s, please fix the syntax and add some context. $\endgroup$
    – Jack D'Aurizio
    Mar 7, 2018 at 20:43
  • $\begingroup$ Mathematica NIntegrate function can be used for plotting as well. For details you can also ask Mathematica Stack Exchange $\endgroup$
    – Yuriy S
    Mar 7, 2018 at 20:46
  • $\begingroup$ I have made the edits $\endgroup$
    – Indeterminate
    Mar 7, 2018 at 20:46
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    $\begingroup$ Though integrating analytically w.r.t. $x$ or $y$ first will make the plotting much faster $\endgroup$
    – Yuriy S
    Mar 7, 2018 at 20:48
  • $\begingroup$ But can NIntegrate integrate the function with a variable V inside the integrand? $\endgroup$
    – Indeterminate
    Mar 7, 2018 at 20:48

1 Answer 1

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As Yuriy mentioned, one may integrate over $x$ fairly easily to get that the double integral is equal to

$$I(V)=\frac{2}{e^V-1} \int_0^{\infty} dy \, \log{\left (\frac{1+e^Ve^{-y^2}}{1+e^{-y^2}} \right )} $$

This is a nice, smooth function that decreases very rapidly and which Mathematica should find simple.

Alternatively, depending on the value of $V$, you can Taylor expand the integrand and express the integral as a sum.

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