# Why does Assuming for Integrate not work as expected?

I'm trying to perform the following integral with Mathematica 7:

Assuming[{t>0},
Integrate[(-s^2 + 4*s*t + 2*alpha*s)^(-(1/2)*D),{s, 0, t}, {alpha, 0, Infinity}]]


If I perform the integration with just the t>0 assumption, Mathematica gives:

(1/(-2 + D))If[Re[D] < 2,
( 2^-D t^(2 - D) (
-8 (-4 + D) Gamma[1 + D/2] Hypergeometric2F1[1 - D/2, D/2, 2 - D/2, 1/4]
+ (-2 + D) D Gamma[D/2] Hypergeometric2F1[2 - D/2, D/2, 3 - D/2, 1/4]
))/((-4 + D) (-2 + D) Gamma[1 + D/2]),
Integrate[(-s (s - 4 t))^(1 - D/2)/s, {s, 0, t}, Assumptions -> Re[D] >= 2 && t > 0]]


which indicates that D must be less than 2. If I now use the additional assumption D<2 with Assuming[{t>0,D<2}, Mathematica says that the integral no longer converges. (Also with exchanged places for the integration variables, the integral can no longer be calculated).

Can anyone solve the mystery?

Thanks, Tobias

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Not an answer to your question, but this will give you a result : Simplify[Integrate[(-s^2 + 4*s*t + 2*alpha*s)^(-(1/2)*bigD), {s, 0, t}, {alpha, 0, Infinity}], Assumptions -> {bigD < 2}]. –  b.gatessucks Jan 22 '13 at 16:00
Please note that D is a reserved word –  Sjoerd C. de Vries Jan 22 '13 at 16:05
Thanks, it looks like assumptions work differently when used directly in Integrate or with Assuming[]. –  Tobias Jan 22 '13 at 18:09

It really diverges.

I'll start with integrating over s. You actually integrate over alpha, so if this is a case where Fubini's theorem does not hold then this analysis might be wrong.

Integrate[(-s^2 + 4*s*t + 2*alpha*s)^(-(1/2)*d), {s, 0, t},
Assumptions -> {t > 0, alpha > 0}]

(* ConditionalExpression[(2 alpha + 4 t)^(1 - d)
Beta[t/(2 alpha + 4 t), 1 - d/2, 1 - d/2], Re[d] < 2] *)


This shows (as you knew) that we require d<2. So let's see where that takes us.

i1 = Integrate[(-s^2 + 4*s*t + 2*alpha*s)^(-(1/2)*d), {s, 0, t},
Assumptions -> {t > 0, d < 2, alpha > 0}];


We can check behavior at the endpoints to assess convergence.

sinf =
Normal[Series[i1, {alpha, Infinity, 2},
Assumptions -> {t > 0, d < 2, alpha > 0}]]

(* Out[102]= (alpha/t)^(d/2) (alpha + 2 t)^(
1 - d) (-((2^(1 - d/2) t)/(alpha (-2 + d))) - (
2^(-1 - d/2) (-16 + 5 d) t^2)/(alpha^2 (-4 + d))) *)


If you look closely you will see that the lead term in alpha is of magnitude alpha^(d/2)*alpha^(1-d)*alpha^(-1). This comes to alpha^(-d). Since d<2 this means we won't have convergence at infinity.

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Thanks for your explanation. This was not my main question though. With b.gatessucks' Simplify-Assumptions I get a result, if I use these assumptions for Integrate itself or use them with Assuming I get the no-convergence error. This is also interesting, since when one performs the integrations one by one starting with s first, it only converges for d<2, starting with alpha, it only converges for d>2. Still mathematica gives a result for d<2 with the Simplify-Assumptions method. –  Tobias Jan 22 '13 at 17:48
This might indeed be a case where Fubinis theorem does not hold, though the s-integral appears here in my problem "naturally" as the outer one (understood as iterated integrals). I will have to think about this more. –  Tobias Jan 22 '13 at 18:08
There is a subtlety that account for getting a result with assumptions. It is that multiple integrals do very little to assess convergence other than in the final integration. If you instead do Simplify[Integrate[(-s^2 + 4*st + 2*alphas)^(-(1/2)*bigD), {s, 0, t}, {alpha, 0, Infinity}, Assumptions -> {bigD < 2}, GenerateConditions -> True], Assumptions -> {bigD < 2}] you will again get a divergence. –  Daniel Lichtblau Jan 22 '13 at 20:44