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Analytically evaluate an integral in quadratic and higher powers?

Arny Toynbee
Posted 8 years ago

Assuming that p1, p4 > 0, are there any substitutions which could get the integration evaluated analytically?

$Assumptions = {p1 > 0 && p4 > 0}
Integrate[1/((p1 + p4)^2*(1 + p4^2)^2), {p4, -Infinity, Infinity}]

If the expression is split up, only one of the three groups evaluates to an analytic result, but the others won't. 

Apart[1/((p1 + p4)^2*(1 + p4^2)^2)]
(* (3*p1^2 - 4*p1*p4 - 1)/((p1^2 + 1)^3*(p4^2 + 1)) + (p1^2 - 2*p1*p4 - 1)/((p1^2 + 1)^2*(p4^2 + 1)^2) + 
  (4*p1)/((p1^2 + 1)^3*(p1 + p4)) + 1/((p1^2 + 1)^2*(p1 + p4)^2)*)

Integrate[(-1 + p1^2 - 2*p1*p4)/((1 + p1^2)^2*(1 + p4^2)^2), {p4, -Infinity, Infinity}]
(* (\[Pi] (p1^2-1))/(2 (p1^2+1)^2) *)
POSTED BY: Arny Toynbee
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Your understanding of Jim Baldwin's argument appears to be incomplete. His argument is taken directly from the definition of an improper integral, and the plot was merely for illustrative purposes and is not central to the argument. The argument shows your integral diverges to positive infinity.

In first-year calculus, it is shown that the integral of a function over an interval containing a pole of order $p \ge 1$ diverges. Your function has a pole of order $2$ at p4 == -p2. Hence it diverges.

Even if you were to consider the Cauchy principal value, your function is positive. Hence the integral diverges to $+\infty$.
POSTED BY: Michael Rogers
Arny Toynbee
Arny Toynbee
Posted 8 years ago

Thanks for your reply Jim Baldwin. Not that for the function 1/x, at x=0, the plot of the function would show that the value of 1/x goes to infinity when approached from both sides. That does not imply that the function 1/x is not integrable!! Similarly, it was pointed out in one of the posts that there is a singularity at p1==-p4, as reflected in your plot (where p1=2).
POSTED BY: Arny Toynbee
Jim Baldwin
Jim Baldwin, Retired
Posted 8 years ago
POSTED BY: Jim Baldwin
POSTED BY: Michael Rogers
Arny Toynbee
Arny Toynbee
Posted 8 years ago

Maple 2017, Maple 7 and Maple 9 all appear to yield the same answer, if you use the commands shown.
POSTED BY: Arny Toynbee
Arny Toynbee
Arny Toynbee
Posted 8 years ago
POSTED BY: Arny Toynbee

Maple says:

enter image description here

enter image description here
POSTED BY: Mariusz Iwaniuk
Arny Toynbee
Arny Toynbee
Posted 8 years ago

Hi Mariusz Iwaniuk, Thanks for your reply. Note however, that both p1>0, p4 >0. The integral is clearly convergent over p4. For e.g. if p1=2, 

NIntegrate[1/((2 + p4)^2*(1 + p4^2)^2), {p4, -Infinity, Infinity}]
(*491.433 *)

There is a singularity when p1==-p4, if p1, p4 are only reals, and not necessarily >0. But it's that which is tractable?

POSTED BY: Arny Toynbee

If assume as an example p1=2 then:

f[x_] := 1/((2 + x)^2 (1 + x^2)^2);
g[e_] := NIntegrate[f[x], {x, -e, e}, Method -> "Trapezoidal"];
Table[{10^e, g[10^e]}, {e, 1, 20}] // Quiet // MatrixForm

enter image description here

and what are the conclusions ?
POSTED BY: Mariusz Iwaniuk

If $p1>0$

enter image description here

then integral does not converge and value is $\infty$

if $\Im(\text{p1})<0$

enter image description here

Reagards,MI
POSTED BY: Mariusz Iwaniuk
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