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 November 2nd, 2019, 12:16 PM #1 Senior Member     Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 169 Thanks: 65 Math Focus: Area of Circle Generalization of Harmonic Series $h(x)$ is a generalized $H_n$, and $H_n = \sum\limits_{k=1}^n \dfrac{1}{k}$. Assumptions:For $n \in \mathbb{N}$, $h(n)=H_n$; $h(1)=1$; $h(x)=h(x-1) + \dfrac{1}{x}$. I started by $$\Large{f(n)= \int_1^n h(z)dz}$$ and at next page I derived at $$\Large{\int_1^n h(z) dz = \ln (n!) + (n-1)k},$$ where $$\large{k=\int_0^1 h(z)dz},$$ which is going to be called Euler-Mascheroni constant. Now, my brain stopped working, so what should be done after $$\Large{\int_1^n h(z) dz = \ln (n!) + (n-1)k}.$$ Thanks from idontknow
 November 2nd, 2019, 12:32 PM #2 Senior Member     Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 169 Thanks: 65 Math Focus: Area of Circle I can just assume $\int h(z) dz = \ln (z!) + (z-1)k$ and check if it holds true for $\int_1^n h(z) dz$.
 November 2nd, 2019, 07:41 PM #3 Senior Member     Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 169 Thanks: 65 Math Focus: Area of Circle How can I show $$\large{\int_0^1 h(z)dz}$$ is convergent and between 0 and 1, or it is equal to $$\large{\lim\limits_{n \rightarrow \infty} H_n - \ln (n)}$$?
November 3rd, 2019, 02:12 PM   #4
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Quote:
 Originally Posted by tahirimanov19 How can I show $$\large{\int_0^1 h(z)dz}$$ is convergent and between 0 and 1, or it is equal to $$\large{\lim\limits_{n \rightarrow \infty} H_n - \ln (n)}$$?
Knowing that $$\Large{h(z) = \int_{0}^{1} \dfrac{1-x^z}{1-z}},$$

I found,
$$\Large{\int_{0}^1 h(z) dz = \int_0^1\left(\frac1{\ln x}+\frac{1}{1-x}\right)\ dx}.$$

What to do next???

 November 3rd, 2019, 03:37 PM #5 Global Moderator   Joined: Dec 2006 Posts: 21,107 Thanks: 2324 Knowing that what??
 November 3rd, 2019, 05:46 PM #6 Senior Member     Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 169 Thanks: 65 Math Focus: Area of Circle Harmonic series...
November 3rd, 2019, 05:48 PM   #7
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Joined: Mar 2015
From: Universe 2.71828i3.14159

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Math Focus: Area of Circle
Quote:
 Originally Posted by tahirimanov19 Knowing that $$\Large{h(z) = \int_{0}^{1} \dfrac{1-x^z}{1-x}} dx,$$ I found, $$\Large{\int_{0}^1 h(z) dz = \int_0^1\left(\frac1{\ln x}+\frac{1}{1-x}\right)\ dx}.$$ What to do next???
I made a mistake... It is $$\Large{h(z) = \int_{0}^{1} \dfrac{1-x^z}{1-x}} dx,$$

 November 4th, 2019, 02:31 AM #8 Senior Member   Joined: Dec 2015 From: Earth Posts: 823 Thanks: 113 Math Focus: Elementary Math $\displaystyle H_n =\ln(n) +\gamma +\epsilon_n .$ All of integrals are correct , the last one maybe has to do with the unknown Last edited by idontknow; November 4th, 2019 at 02:33 AM.

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