Average wavelength for blackbody radiation

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jimjohnson
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The maximum wavelength for blackboby radiation is .29/Temperature. Is the average wavelength 1.84 times the maximum for all temperatures?
 
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As shown in both ref., the wavelength where the emission intensity is maximum is .29/T. But I want to know the avg wavelength at that temperatue. So I think the wavelength where the area under the intensity ghaph is equal on both sides would be the average, but I do not know how to calculate it from gragh equation. My estimate working back from other calculations was 1.84 times the max wavelength.
 
You can calculate the following:

Use Planck's Law for the blackbody radiation power as a function of T and λ

http://en.wikipedia.org/wiki/Planck's_law

u(T,λ) = [2hc25]/[e(hc/λkT) - 1]

The average wavelength is then given by:

λavg = ∫λ u(T,λ) dλ / ∫ u(T,λ) dλ

where both integrals are from 0 to ∞. The appendix of the above wiki solves the important integrals.

Bob S
 
Do you know if the average is 1.84 the maximum for all temperatures? I can not do the math. The following quote from the source may relate: This function peaks for hc = 4.97λkT, a factor of 1.76 shorter in wavelength (higher in frequency) than the frequency peak. Thanks jimj