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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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  • $\begingroup$ Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the faq! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! $\endgroup$ – user9660 Mar 2 '15 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 Mar 2 '15 at 10:50
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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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