# Allan Variance integral for a first order Markov process

I am using Mathematica to solve this definite integral:

4*(q*T)^2*Integrate[(Sin[Pi*f*t])^4/(Pi*f*t)^2/(1 + (2*Pi*f*T)^2),
{f, 0, Infinity}]


$$T$$ and $$t$$ are real positive. Mathematica is unable to solve this. In IEEE 952-1997, Annex C, equation C.16 the solution to this integral is given as:

$$(q*T)^2/t*[1-(T/2/t)*(3-Exp[-t/T]+Exp[-2*t/T]]$$

Is there a way to get Mathematica to solve this?

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You can add Assumptions to Integrate:

4*(q T)^2 Integrate[(Sin[Pi  f  t])^4/(Pi  f  t)^2/(1 + (2 Pi  f*T)^2),
{f, 0, Infinity}, Assumptions -> T > 0 \[And] t > 0]


which takes a little while to evaluate and gives

(q^2 T^2 (2 t + (-3 + E^(-((2 t)/T)) (-1 + 4 E^(t/T))) T))/(2 t^2)


This isn't quite the same as the answer you give due to the factor 4 E^(t/T) in the Mathematica result instead of E^(t/T) needed to match your answer. Possibly a typo in the integrand -- when I copied the expression you gave, I had to edit it to insert spaces between the variable names.

• This is a bug in Integrate because (q^2 T^2 (2 t + (-3 + E^(-((2 t)/T)) (-1 + 4 E^(t/T))) T))/( 2 t^2) /. {q -> 1., T -> 1, t -> 1} results in 0.168091 whereas NIntegrate[(Sin[ Pi f t])^4/(Pi f t)^2/(1 + (2 Pi f*T)^2) /. {q -> 1, T -> 1, t -> 1}, {f, 0, Infinity}] produces 0.0420228. Commented yesterday
• Is this a typo instead of a bug: your NIntegrate expression is missing the factor 4*(q T)^2 that's outside the integral in the question. That factor equals 4 for your example of q=T=1
• tad (@does not work): You are right. Taking into account that multiplier, both results are equal to 0.168091. Sorry for the trouble. Commented yesterday