How to work out the parameter in a definite integration which has an exact value while the integration doesn't have an analytical solution?

Here is the equation I'm trying to solve:

NIntegrate[1/(E^(1/(λ T)) - 1), {λ, 200, 220}] == 1000


T is the parameter I'm trying to work out. I tried to solve it with InverseFunction[] but it seemed not to work:

te[ii_] := InverseFunction[NIntegrate[1/(-1 + E^(1/(# λ))), {λ, 200, 220}] &][ii]
Evaluate[te[1000]]


What's wrong with my solution? How to correct it?

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Here's a slight improvement of b.gatessucks's answer, adding a (mostly effective) initial guess:

te[ii_?NumericQ, opts___] := (\[FormalCapitalT] /. First@FindRoot[
NIntegrate[1/(E^(1/(λ \[FormalCapitalT])) - 1), {λ, 200, 220}] == ii,
{\[FormalCapitalT], (0.5437727672315316 + ii (0.5440978395984463 +
ii (0.12244298721801757 + (0.009523809523809525 +
0.0002380952380952381 ii) ii)))/(476.1910924718744 +
ii (247.61910924718742 + ii (30. + ii)))},
Evaluated -> False, opts])


Here's how I obtained the initial guess:

PadeApproximant[InverseSeries[
Integrate[#, {λ, 200, 220}] & /@
Normal[Series[1/(E^(1/(λ ii)) - 1), {ii, Infinity, 7}]] + O[ii, Infinity]^8],
{ii, Infinity, 4}] // N // FullSimplify

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Wow, what a advanced solution…Er, I wonder where I can find the introduction of Evaluated, I can't find it in the help, and, it seems that PadeApproximant here isn't necessary, why you choose it? – xzczd Aug 1 '12 at 11:42
The Evaluated option is indeed undocumented, but it has been discussed a lot on this site; search around. As for PadeApproximant[], it's for generating a rational function approximation; they tend to be more accurate than truncations of Taylor polynomials for the same amount of arithmetic operations. – J. M. Aug 1 '12 at 12:05
For example this? OK, it's very interesting, now…I'm sorry! But I still can't understand the independent variable opts___ which I thought I would make it out after I knew the meaning of Evaluated… – xzczd Aug 1 '12 at 13:07
The opts is for the case where you need to pass options to FindRoot[]. Now that I think about it, I should have also made provisions for passing options to NIntegrate[]... maybe later. – J. M. Aug 1 '12 at 13:19
So, here I can just remove it? (Trying…) ………er, it seems that the ?NumericQ and Evaluated(whether it's True or False) are both unnecessary now… – xzczd Aug 1 '12 at 13:36

You can use FindRoot :

func[T_?NumericQ] :=  NIntegrate[1/(E^(1/(\[Lambda] T)) - 1), {\[Lambda], 200, 220}]

sol = FindRoot[func[T] == 1000, {T, 0.1}]

(* {T -> 0.240468} *)

func[sol[[1, 2]]]

(* 1000. *)


FindRoot needs an initial guess for the unknown; you can get a rough estimate by plotting func for instance.

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Maybe for this case the initial guess isn't that important since the te[ii] is actually monotonous? – xzczd Aug 1 '12 at 13:12
@xzczd I agree. – b.gatessucks Aug 1 '12 at 13:24