# How can I obtain a regularized value of this integral using Mathematica?

I want to know the regularized value of this integral. Wolfram Mathematica fails. $$\int_0^\infty \psi'(x+1)dx$$

I have two conjectures, it is either $$\gamma$$ or $$0$$.

I attempted

Sum[f[s x],{x,1,Infinity},Regularization->"Borel"]//FullSimplify

Limit[s %,s→0]


with all available regularization methods instead of "Borel" but Mathematica produced no result.

• What is psi, what is gamma, what is prime. When you ask a question you really should define your terms. Nov 8, 2019 at 18:40
• ... also, which regularization in particular? the result is non-unique. Nov 8, 2019 at 19:53

HoldForm[PolyGamma[1, 1 + x] == Sum[1/(1 + k + x)^2, {k, 0, Infinity}]] // TeXForm


$$\psi ^{(1)}(1+x)=\sum _{k=0}^{\infty } \frac{1}{(1+k+x)^2}$$ Then:

INT = Integrate[1/(1 + k + x)^2, {x, 0, Infinity}]
(* ConditionalExpression[1/(1 + k), Im[k] != 0 || Re[k] >= -1] *)

INT[[1]]
(* 1/(1 + k) *)

Sum[INT[[1]], {k, 0, Infinity}, Regularization -> "Borel"]
(* EulerGamma *)


Another way:

$$\int_0^{\infty } \psi ^{(1)}(1+x) \, dx=\int_0^{\infty } \left(\int_0^{\infty } \frac{e^{-t (1+x)} t}{1-e^{-t}} \, dt\right) \, dx=\int_0^{\infty } \left(\int_0^{\infty } \frac{e^{-t (1+x)} t}{1-e^{-t}} \, dx\right) \, dt=\int_0^{\infty } \frac{1}{-1+e^t} \, dt=\int_0^{\infty } \left(\sum _{j=1}^{\infty } \exp (-j t)\right) \, dt=\sum _{j=1}^{\infty } \int_0^{\infty } \exp (-j t) \, dt=\sum _{j=1}^{\infty } \frac{1}{j}=\gamma$$

Last sum with Borel regularization.

• I wonder what does [[1]] mean and what elements are necessary (I think texform is not?) Nov 8, 2019 at 20:00
• @Anixx It means [[1]] extract form ConditionalExpression a 1/(1 + k). Nov 8, 2019 at 20:11
• So, the answer is $\gamma$ to my disappointment Nov 8, 2019 at 20:15
• @Anixx It seems so. Nov 8, 2019 at 20:36
• I suspected this because $$\int_0^\infty \psi'(x+1)dx-\gamma=\psi(\infty)=\ln(\infty)=\int_1^\infty \frac1xdx=\sum_{k=1}^\infty \frac1k-\gamma=0$$ (the concept of divergent limits at infinity would be infinitely useful) Nov 8, 2019 at 20:43