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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.

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

1 Answer 1

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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.

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  • $\begingroup$ I wonder what does [[1]] mean and what elements are necessary (I think texform is not?) $\endgroup$
    – Anixx
    Nov 8, 2019 at 20:00
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    $\begingroup$ @Anixx It means [[1]] extract form ConditionalExpression a 1/(1 + k). $\endgroup$ Nov 8, 2019 at 20:11
  • $\begingroup$ So, the answer is $\gamma$ to my disappointment $\endgroup$
    – Anixx
    Nov 8, 2019 at 20:15
  • $\begingroup$ @Anixx It seems so. $\endgroup$ Nov 8, 2019 at 20:36
  • $\begingroup$ 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) $\endgroup$
    – Anixx
    Nov 8, 2019 at 20:43

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