# Evaluation of integral with parameters

I am trying to solve the following integral having two strictly positive parameters:

Integrate[(u - 1)^2 * (1 - u^(-1/xi))^(k - 3) * u^(-3/xi - 3),
{u, 1, Infinity}, Assumptions -> {xi > 0, k > 0}]


but the returned solution only holds for $$k > 2$$.

If I integrate the same expression by making separate assumptions on $$k$$ (that is, first I integrate assuming $$k > 2$$, then I integrate assuming $$k = 2$$ and so on ...), I am able to get a solution for $$k=1$$, $$k=2$$ and $$k>2$$. But when I run

Integrate[(u - 1)^2 * (1 - u^(-1/xi))^(k - 3) * u^(-3/xi - 3),
{u, 1, Infinity}, Assumptions -> {xi > 0, 0 < k < 1}]


Mathematica returns the same unevaluated expression. The same happens when I integrate the expression assuming $$1.

Additional attempts: I tried to solve the integral by first applying the function FullSimplify but nothing changes. I also developed the term (u - 1)^2 and multiplied each of its element for (1 - u^(-1/xi))^(k - 3) * u^(-3/xi - 3) to find out that the integrals do not converge on {1, Infinity} for 0<k<1 nor for 1<k<2.

How can I solve the integral for $$0 and $$1?

Let us investigate the behavior of the integrand as $$u\to 1$$

Series[(u - 1)^2*(1 - u^(-1/xi))^(k - 3)*u^(-3/xi - 3), {u, 1, 2},
Assumptions -> {xi > 0, k > 0}] // Normal


(-1 + u)^(-1 + k) xi^(3 - k)

and as $$u\to \infty$$

Series[(u - 1)^2*(1 - u^(-1/xi))^(k - 3)*u^(-3/xi - 3), {u, Infinity,   2},
Assumptions -> {xi > 0, k > 0}] // Normal


u^(-3 - 3/xi) (1 - 2 u + u^2) (1 - u^(-1/xi))^(-3 + k)

The above results prove the convergence of the integral under consideration if xi > 0, k > 0.

In view of it we may consider

j = Integrate[(u - 1)^2*(1 - u^(-1/xi))^(k - 3)*u^(-3/xi - 3), {u, 1,  Infinity},
Assumptions -> {xi > 0, k > 0},  GenerateConditions -> False]


xi (2/(k (2 - 3 k + k^2)) + Gamma[-2 + k] (-((2 Gamma[3 + xi])/Gamma[1 + k + xi]) + Gamma[3 + 2 xi]/Gamma[1 + k + 2 xi]))

Numeric calculations

N[j /. {k -> 1/2, xi -> 1/10}]


0.00127714

and

NIntegrate[(u - 1)^2*(1 - u^(-1/xi))^(k - 3)* u^(-3/xi - 3) /. {k -> 1/2, xi -> 1/10}, {u, 1, Infinity}]


0.00127714

confirm it.

• It should be noticed that Limit[j, k -> 1, Assumptions -> xi > 0] produces xi (-((2 Gamma[3 + xi] (-1 + EulerGamma + PolyGamma[0, 2 + xi]))/ Gamma[2 + xi]) + (Gamma[ 3 + 2 xi] (-1 + EulerGamma + PolyGamma[0, 2 + 2 xi]))/ Gamma[2 + 2 xi]). Jun 19 at 18:45
• What about integer k<2 ? Gamma[] is ComplexInfinity on negative integers, so the Gamma[-2 + k] part would be problematic for integer k<2. I suppose it does not matter too much though since k can take on any other non-integer value
– ydd
Jun 19 at 19:48
• Thanks @user64494 for your help. If you showed that the integral converges for $xi>0, k>0$, why does Mathematica return the same solution only for $k>2$ when GenerateConditions is put equal to True? In addition, if I need to evaluate j in $k==1$ I can't because of the indeterminate form. Is it correct in this case to use the value returned by Integrate[(u - 1)^2 * (1 - u^(-1/xi))^(k - 3) * u^(-3/xi - 3), {u, 1, Infinity}, Assumptions -> {xi > 0, k == 1}] in order to evaluate the solution in $k==1$?
– cpcf
Jun 19 at 22:04
• Try taking limits as k aapraches 1 or 2. It looks like there might be singular terms that cancel. Jun 19 at 22:21