Infinite Integrations?!?!

Let

$\displaystyle a_k= I_{k,n}(x) = \psi^{n-k} (x^{k}\ln x)\;\;\;\; \text{for} \;\;\; n,k \in N \; , k\leq n$

Where we have : $\displaystyle I_{n,n}(x) = x^{n}\ln x \;\;\; \text{and} \;\;\;\; I_{0,n}(x) = \psi^{n}(\ln x )$

By definition:

$\displaystyle a_k = \psi^{n-k-1} (\psi (x^{k}\ln x))$

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Working on $\psi (x^{k}\ln x)$ , integrate by parts with $u=\ln x , dv=x^{k}dx$ :

$\displaystyle \psi (x^{k}\ln x) = \frac{x^{k+1}}{k+1}\ln x - \psi ( \frac{x^{k}}{k+1} )$

$\displaystyle = \frac{x^{k+1}}{k+1}\ln x - \frac{x^{k+1}}{(k+1)^2}$

Therefore:

$\displaystyle a_k = \psi^{n-k-1} ( \frac{x^{k+1}}{k+1}\ln x - \frac{x^{k+1}}{(k+1)^2} )$

$\displaystyle = \frac{\psi^{n-(k+1)} (x^{k+1}\ln x )}{k+1} - \frac{ \psi^{n-k-1} (x^{k+1}) }{ (k+1)^2}$

$\displaystyle = \frac{a_{k+1}}{k+1} - \frac{ {\color{#D61F06}{\psi^{n-k-1} (x^{k+1})}} }{ (k+1)^2}$

$\mathbf{Preposition :}$

$\displaystyle \psi^a (x^b) = \frac{b!}{(a+b)!}x^{a+b} , \;\; \forall a,b \in N$

Can be proved easily by induction on $a$ . Now substitute $a= n-k-1 \; ,\; b= k+1$ :

$\displaystyle {\color{#D61F06}{\psi^{n-k-1} (x^{k+1})}} = \frac{x^{n}}{n!} (k+1 )!$

Hence:

$\displaystyle a_k= \frac{a_{k+1}}{k+1} - \frac{x^{n}k!}{n! (k+1)}$

Now replace $k$ by $k-1$ then solve rearrange for $a_k$ :

$\displaystyle a_k = (k)a_{k-1} + \frac{x^{n}(k-1)!}{n!}$

Now solve this recursion by the following fact:

$\displaystyle a_k= M_k + B_ka_{k-1} <=> a_k = \Bigg(\prod_{i=1}^k B_i \Bigg)\Bigg(a_0 + \sum_{j=1}^k \frac{M_j}{\prod_{m=1}^j B_m} \Bigg)$

We obtain at last :

$\displaystyle a_k = k!( a_0 + \frac{x^{n}}{n!} H_k ) \;\;\; \text{where}\; H_k \; \text{is the}\; k^{th} \; \text{Harmonic number}$

Now let $k=n$ and solve for $a_0$ :

$\displaystyle a_0 = \frac{a_n - x^nH_n}{n!}$

From the definition of $a_k$ :

$\displaystyle a_0 = \psi^{n}(\ln x ) \;\;\;\;\; \text{and} \;\;\;\;\; a_n = x^n\ln x$

Hence:

$\displaystyle \boxed{ \psi^{n}(\ln x ) = \frac{x^n}{n!} (\ln x - H_n ) }$

Then :

$\displaystyle \phi (n) = \psi^{n}(\ln x )|_{x=1+\frac{1}{n} } = \frac{\bigg(1+\frac{1}{n} \bigg)^n}{n!} (\ln \bigg(1+\frac{1}{n} \bigg) - H_n )$

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$\displaystyle => L= \lim_{n\to \infty} [ n!\phi (n) +e\ln n ]$

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$\displaystyle = \lim_{n\to \infty} [\bigg(1+\frac{1}{n} \bigg)^n ( \ln \bigg(1+\frac{1}{n} \bigg) - H_n ) +e\ln n ]$

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$\displaystyle = \lim_{n\to \infty} [(e)(0-H_n) +e\ln n ]$

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$\displaystyle = -e\lim_{n\to \infty} ( H_n - \ln n )$

$$

By the definition of the Euler-Mascheroni Constant :

$\displaystyle L= -e\gamma$

So our desired answer is:

$\displaystyle e+L =\boxed{ e(1-\gamma)}$