# Showing that $\gamma = -\int_0^{\infty} e^{-t} \log t \,dt$, where $\gamma$ is the Euler-Mascheroni constant.

I'm trying to show that

$$\lim_{n \to \infty} \left[\sum_{k=1}^{n} \frac{1}{k} - \log n\right] = -\int_0^{\infty} e^{-t} \log t \,dt.$$

In other words, I'm trying to show that the above definitions of the Euler-Mascheroni constant $\gamma$ are equivalent.

In another post here (which I can't seem to find now) someone noted that

$$\int_0^{\infty} e^{-t} \log t \,dt = \left.\frac{d}{dx} \int_0^{\infty} t^x e^{-t} \,dt \right|_{x=0} = \Gamma'(1) = \psi(1),$$

where $\psi$ is the digamma function. This may be a good place to start on the right-hand side.

For the left-hand side I was tempted to represent the terms with integrals. It is not hard to show that

$$\sum_{k=1}^{n} \frac{1}{k} = \int_0^1 \frac{1-x^n}{1-x} \,dx,$$

but I'm not sure this gets us anywhere.

Any help would be greatly appreciated.

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It is easy to prove that the function

$$f_n(x) = \begin{cases} \left( 1 - \frac{x}{n}\right)^n & 0 \leq x \leq n \\ 0 & x > n \end{cases}$$

satisfies $0 \leq f_n(x) \uparrow e^{-x}$. Thus by dominated convergence theorem,

$$\int_{0}^{\infty} e^{-x} \log x \; dx = \lim_{n\to\infty} \int_{0}^{n} \left( 1 - \frac{x}{n}\right)^n \log x \; dx.$$

Now by the substitution $x = nu$, we have

\begin{align*} \int_{0}^{n} \left( 1 - \frac{x}{n}\right)^n \log x \; dx &= n\int_{0}^{1} \left( 1 - u\right)^n (\log n + \log u) \; du \\ &= \frac{n}{n+1}\log n + n\int_{0}^{1} \left( 1 - u\right)^n \log u \; du \\ &= \frac{n}{n+1}\log n + n\int_{0}^{1} v^n \log (1-v) \; dv \\ &= \frac{n}{n+1}\log n - n\int_{0}^{1} v^n \left( \sum_{k=1}^{\infty} \frac{v^k}{k} \right) \; dv \\ &= \frac{n}{n+1}\log n - n \sum_{k=1}^{\infty} \frac{1}{k(n+k+1)} \\ &= \frac{n}{n+1}\log n - \frac{n}{n+1} \sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{n+k+1}\right) \\ &= \frac{n}{n+1} \left( \log n - \sum_{k=1}^{n+1} \frac{1}{k} \right). \end{align*}

Therefore taking $n \to \infty$ yields $-\gamma$. If you are not comfortable with the interchange of integral and summation, you may perform integration by parts as follows:

\begin{align*} \int_{0}^{1} v^n \log (1-v) \; dv &= \left. \frac{v^{n+1} - 1}{n+1} \log (1-v) \right|_{0}^{1} - \int_{0}^{1} \frac{v^{n+1} - 1}{n+1} \cdot \frac{1}{v - 1} \; dv \\ &= - \frac{1}{n+1} \int_{0}^{1} \frac{1 - v^{n+1}}{1 - v} \; dv \end{align*}

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Thank you. That's a nice trick at the beginning there. –  Antonio Vargas Feb 23 '12 at 5:32

Since

$$\frac{{\Gamma '\left( x \right)}}{{\Gamma \left( x \right)}} = - \gamma - \frac{1}{x} + \sum\limits_{v = 1}^\infty {\frac{x}{{v\left( {x + v} \right)}}}$$

We evaluate the expression at $x=1$ to get

$$\frac{{\Gamma '\left( 1 \right)}}{{\Gamma \left( 1 \right)}} = \Gamma '\left( 1 \right) = - \gamma - 1 + \sum\limits_{v = 1}^\infty {\frac{1}{{v\left( {1 + v} \right)}}}$$

But since $$\sum\limits_{v = 1}^\infty {\frac{1}{{v\left( {1 + v} \right)}}}=1$$

we get

$$\Gamma '\left( 1 \right) = - \gamma$$

This would be an instant consequence of the proof that the digamma function is defined by

$$\psi \left( x \right) = \frac{{\Gamma '\left( x \right)}}{{\Gamma \left( x \right)}} = - \gamma - \frac{1}{x} + \sum\limits_{v = 1}^\infty {\frac{x}{{v\left( {x + v} \right)}}}$$

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