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These are the steps in order to get the Euler-Mascheroni constant $\gamma$ there were steps before where I'm starting but I didn't have questions about the justifications of those so I didn't include them.

$\int^1_0\ln(x)d(e^{-x}-1)+\int^\infty_1\ln(x)de^{-x}$

(1)=$\int^1_0\frac{1-e^{-x}}{x}dx-\int^\infty_1\frac{e^{-x}}{x}dx$

(2) $=\lim_{n\rightarrow\infty}[\int^1_0 \frac{1-(1-\frac{x}{n})^n}{x}dx-\int^n_1\frac{(1-\frac{x}{n})^n}{x}dx]$

(3)$=\lim_{n\rightarrow\infty}[\int^\frac{1}{n}_0\frac{1-(1-t)^n}{t}dt-\int^1_\frac{1}{n}\frac{(1-t)^n}{t}dt]$

(4)$=\lim_{n\rightarrow\infty}[\int^\frac{1}{n}_0\frac{1-(1-t)^n}{t}dt+\int^1_\frac{1}{n}[\frac{1-(1-t)^n}{t}-\frac{1}{t}]]dt$

(5)$=\lim_{n\rightarrow\infty}[\int^1_0\frac{1-(1-t)^n}{t}dt-\int^1_{\frac{1}{n}}\frac{1}{t}dt]$

(6) $=\lim_{n\rightarrow\infty}[\int^1_0\frac{1-t^n}{1-t}dt+ln(\frac{1}{n})]$

(7) $=\lim_{n\rightarrow\infty}[\int^1_0(t^{n-1}+t^{n-2}+...+t+1)dt+\ln(\frac{1}{n})]$

(8)$=\lim_{n\rightarrow\infty}(\frac{1}{n}+\frac{1}{n-1}+...+\frac{1}{2}+1-\ln(n)):=\gamma$

My rational:

(1) I don't understand this at all. Somehow, two sides that were identical are different. It's upsetting.

(2)$\lim_{n\rightarrow\infty}(1-\frac{x}{n})^n=e^{-x}$ so it's just a substitution.

(3) I don't like this one either. It doesn't seem like $\frac{x}{n}=t$ and $x=t$ is possible like it looks here. Even with the change of the integration domain.

(4) Just an arithmetic thing. No worries here.

(5) A change of domain incurred in the first part but the inside of the integral did not change. I suppose it's because you moved part of the second part over but I don't understand how in the world you would do that.

(6)The second part looks golden. The first part is not the same so I don't get that.

(7) We're just writing the first part out. This is good.

(8) And I'm back to not understanding.

Any help is appreciated!

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  • $\begingroup$ A related question. $\endgroup$ – Lucian Mar 18 '15 at 13:35
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(1) is integration by parts, $$ \int_0^1 \log{x} d(e^x-1) = [(e^x-1)\log{x}]_0^1 - \int_0^1 \frac{e^x-1}{x} \, dx. $$ The boundary term evaluates to $0$, which can be seen by taking the bottom limit to be $a$ and taking $a \downarrow 0$. (The other integral works in the same way.)

(2) requires checking that the errors tend to $0$, which is going to be a fair amount of work, but not intractable.

(3) $x$ is just the integration variable, and valid substitution like is carried out here doesn't change the value of the integral, and so isn't going to change the limit.

(5) This is $$ \int_a^b f(t) \, dt + \int_b^c f(t) \, dt = \int_a^c f(t) \, dt, $$ with $a=0$. $b=1/n$, $c=1$ and $f(t) = (1-(1-t)^n)/t$.

(6) The change of variables $u=1-t$ has been executed in the first integral (but $u$ is then relabelled to $t$, which does look confusing).

(8) Swapping the sum and the integral, we get $$ \int_0^1 (t^{n-1}+t^{n-2}+\dotsb+t+1) \, dt = \sum_{k=0}^{n-1} \int_0^1 t^{k} \, dt, $$ and you know that $$ \int_0^1 t^k \, dt = \frac{1}{k+1}, $$ so the sum is $$ \frac{1}{n-1+1} + \dotsb + \frac{1}{1+1} + \frac{1}{0+1} = \frac{1}{n} + \frac{1}{n-1} + \dotsb + \frac{1}{2} + 1 $$

(Say if you want some more detail)

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  • $\begingroup$ What do you mean be errors in 2? Is that only for the second piece of the equation? $\endgroup$ – chriskinda Mar 18 '15 at 14:38
  • $\begingroup$ I mean that there are two bits of the second integral you ignore when you make that substitution: there's a $\int_n^{\infty} e^{-x}/x \, dx $ that's been removed. You have to check that that tends to $0$. The other thing is basically checking that integrating $x^{-1}[e^x-(1+x/n)^n]$ has some sort of bound (so that you can swap $\int \lim$ to $\lim \int$). $\endgroup$ – Chappers Mar 18 '15 at 14:46

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