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I want to know how Mathematica evaluates this integral:

$PV \int _ {0} ^ {i \infty} d \tau \, \frac{e^{-\frac{\tau^2}{2 M^2}}(b^2 - 3 τ^2)^2 (|b^2+\tau^2|-(b^2-\tau^2))}{\tau (\tau^2 - b^2)}$, where the Cauchy Principal Value is taken and $b>0 \, \& \, M>0$.

Mathematica gives me as result

$b^2(-i \pi b^2-\gamma b^2-18 M^2+4b^2e^{-\frac{b^{2}}{2M^2}}(i \pi + Ei(\frac{b^2}{2M^2}))+2 b^2 \log (M)+b^2 \log (2))$

where $\gamma$ is the Euler constant

I am far from sure that the above result is correct, so I would like to know the intermediate steps that Mathematica uses to get this result. Any help (either on the intermediate steps or the validity of the result)?

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    – user9660
    Commented Mar 2, 2015 at 10:20
  • $\begingroup$ @user127054 This integral diverges strongly. Not because of the denominator but of the Exp[-t^2] term which explodes on the imaginary axis like Exp[z^2] for z real. Maybe there is a typing error ... see my answer. $\endgroup$
    – Dr. Wolfgang Hintze
    Commented Mar 2, 2015 at 10:50

1 Answer 1

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First of all correcting the sign in the exponential we get

f2 = Integrate[
  Exp[r^2/(2 M^2)] (b^2 - 3 r^2)^2 (Abs[b^2 + r^2] - (b^2 - r^2))/(
   r (r^2 - b^2)), {r, 0, I \[Infinity]}, 
  Assumptions -> {M > 0, b > 0]

(* Out[278]= 
18 b^2 M^2 + 4 b^4 E^(b^2/(2 M^2)) ExpIntegralEi[-(b^2/(2 M^2))]
*)

Taking the pricipal value has no influence as there is no singularity on the imaginary r-axis.

Maybe this is already the result expected by user127054.

As for a step-by-step approach I would recommend taking the indefinite integral which would give the antiderivative. But in this case MMA does not provide the antiderivative.

More details:

Transforming the integral to the real axis by letting r -> I s we have for the integrand (to be taken between s = 0 and s = +inf)

I Exp[r^2/(2 M^2)] (b^2 - 3 r^2)^2 (Abs[b^2 + r^2] - (b^2 - r^2))/(
   r (r^2 - b^2)) /. r -> I s // Simplify

(*
Out[294]= (E^(-(s^2/(
  2 M^2))) (b^2 + 3 s^2)^2 (b^2 + s^2 - Abs[b^2 - s^2]))/(s (b^2 + s^2))
*)

It seems that there is a singularity at s = 0.

But let us look at the series expansion and observe that b^2 > s^2 since s->0 and b>0:

Simplify[Series[%, {s, 0, 2}] // Normal, b^2 > s^2]

(*
Out[295]= 5 s^3 + b^2 (2 s - s^3/(2 M^2))
*)

Hence there is no singularity at s = 0 and therefore none on the positive s axis. Hence taking the principal has no effect on the result.

Series expansion in b of the result are (for M->1)

at b = 0

Series[f2 /. M -> 1, {b, 0, 4}, Assumptions -> b > 0] // Normal

(*
Out[304]= 18 b^2 + b^4 (4 EulerGamma - 4 Log[2] + 8 Log[b])
*)

at b = inf

Series[f2 /. M -> 1, {b, \[Infinity], 4}, Assumptions -> b > 0] // Normal

(*
Out[306]= 10 b^2 + 16 - 64/b^2 + 384/b^4
*)
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