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How would one go about proving $$\int_{0}^1\frac{e^x-1}{x/2}\ dx=\sum_{n=0}^\infty\frac{1}{\binom{n+2}{2}}\frac{1}{n!}(0!+1!+2!+3!+...+n!)$$

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2 Answers 2

Let $$f(x)=\sum_{n=0}^{\infty} \frac{1} {(n+2)!}(0!+1!+2!+3!+...+n!)x^{n+2}$$ Then we obtain a differential equation $$f'(x)=f(x)-\log(1-x)$$ This solves to $$ f(x)=e^x\int_0^x (-e^{-t}\log (1-t))dt$$ The value that you want to find is $2f(1)$, so plug in $x=1$, $$ f(1)=e\int_0^1 (-e^{-t}\log (1-t))dt$$ A substitution $t$ to $1-t$, we have $$ f(1)=\int_0^1 (-e^t\log t)dt$$ Then integration by part gives $$ f(1)=-\log t (e^t -1)|_0^1+\int_0^1\frac{e^t-1}{t}dt$$ Thus $$2f(1)=\int_0^1\frac{e^t-1}{t/2}dt.$$ This is the desired formula

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There is a little point that I have to make clear. The series has radius of convergence 1, so the argument of "plugging in $x=1$" is actually a use of Abel's theorem. –  i707107 Mar 29 '13 at 1:27

I had a direct approach. However I got stuck at the last step. If someone can check the steps and suggest a way to proceed, I would appreciate it:

Observe that $$\frac{e^x - 1}{x/2} = \frac{2}{x}\sum_{k=1}^\infty \frac{x^k}{k!}$$

Since x is in the nonnegative range, by Monotone convergence theorem, we have

$$\int_{0}^{1}\frac{e^x - 1}{x/2}dx = 2\sum_{k=1}^\infty \int_{0}^{1}\frac{x^{k-1}}{k!}dx$$ $$= 2\sum_{k=1}^\infty \frac{1}{k!k}$$ $$= \sum_{k=0}^\infty \frac{2!}{(k+1)!(k+1)} = \sum_{k=0}^\infty \frac{2!(k+2)}{(k+1)!(k+1)(k+2)}$$ $$= \sum_{k=0}^\infty \frac{2!}{(k+2)!}\frac{k+2}{k+1}$$ $$ = \sum_{k=0}^\infty \frac{1}{\binom{k+2}{2}k!}\frac{k+2}{k+1}$$

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Just a thought, but at x=0, the integrand is undefined. It does approach 1 from RHS. Could that be the issue here? –  Gautam Shenoy Mar 28 '13 at 9:07
    
That is not an issue. The integrand has removable singularity as x approach to 0. –  i707107 Mar 29 '13 at 1:19
    
Yours gives another summation expression of the integral. I don't see how to go further from there. –  i707107 Mar 29 '13 at 1:28
    
When you compare the given RHS with my final expression, what really bothers me is that it seems to imply $\frac{k+2}{k+1} =$ sum of first k factorials!!! I can't seem to pinpoint the error. –  Gautam Shenoy Mar 29 '13 at 8:17
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Not necessarily. For example, $2a+5b+4c=3a+2b+5c$ does not imply $2=3, 5=2, 4=5$. You must deal with the whole sum, not the individual terms. –  i707107 Mar 29 '13 at 8:29

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