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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@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. – Dr. Wolfgang Hintze – 2015-03-02T10:50:10.720