real analysis - Alternative approaches to showing that $gamma=int_0^infty left(frac{1}{1+x^a}-frac{1}{e^x}right),frac1x,dx$, $a>0$

Starting from the limit definition of the Euler-Mascheroni constant $\gamma$ as given by

$$\gamma=\lim_{n\to \infty}\left(-\log(n)+\sum_{k=1}^n\frac1k\right)\tag 1$$

we can show that $\gamma$ has an integral representation

$$\gamma=\int_0^\infty\left(\frac{1}{e^x-1}-\frac{1}{xe^x}\right)\,dx \tag 2$$

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Proof of $(2)$: This is provided for completeness only and one can skip this part without losing context.

To show that the integral in $(2)$ is equivalent to $(1)$, we can proceed as follows.

$$\begin{align} \int_0^\infty\left(\frac{1}{e^x-1}-\frac{1}{xe^x}\right)\,dx&= \int_0^\infty \frac{e^{-x}}{1-e^{-x}}\left(1-\frac{1-e^{-x}}{x}\right)\,dx\\\\ &=\sum_{k=1}^\infty \int_0^\infty\left(e^{-kx}-\frac{e^{-(kx}-e^{-(k+1)x}}{x}\right)\,dx\\\\ &=\sum_{k=1}^\infty \left(\frac{1}{k}-\log\left(\frac{k+1}{k}\right)\right)\\\\ &=\lim_{n\to \infty}\sum_{k=1}^n \left(\frac1k -\log\left(\frac{k+1}{k}\right)\right)\\\\ &=\lim_{n\to \infty}\left(-\log(n+1)-\sum_{k=1}^n\frac1k\right)\\\\ &=\lim_{n\to \infty}\left(-\log(n)-\sum_{k=1}^n\frac1k\right)\\\\ \end{align}$$

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Another integral representation for $\gamma$ is given by

$$\gamma=\int_0^\infty \left(\frac{1}{x(1+x^a)}-\frac{1}{xe^x}\right) \,dx \tag 3$$

for $a>0$.

Equipped with $(2)$, we can show the equivalence of $(3)$ with $(1)$ by showing that

$$\int_0^\infty \left(\frac{1}{x(1+x^a)}-\frac{1}{e^x-1}\right)\,dx=0\tag 4$$

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To prove $(4)$, I proceeded as follows.

$$\begin{align} \lim_{\epsilon\to 0}\int_{\epsilon}^\infty\left(\frac{1}{x(1+x^a)}-\frac{1}{e^x-1}\right)\,dx &=\lim_{\epsilon\to 0}\left.\left(-\frac1a \log(1+x^{-a})+\log(1-e^{-x})\right)\right|_{\epsilon}^{\infty}\\\\ &=\lim_{\epsilon\to 0}\left(\frac1a \log(1+\epsilon^{-a})+\log(1-e^{-\epsilon})\right)\\\\ &=0 \end{align}$$

And we are done!

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This approach seemed a bit cumbersome and indirect.

QUESTION: So, what are alternative approaches to establishing equivalence of $(3)$ and $(1)$?

Answer

Integrating by parts, $$ \int_0^{\infty} \left( \frac{1}{1+x^a} - e^{-x} \right) \frac{dx}{x} = 0-0 + \int_0^{\infty} \left( \frac{ax^a}{x(1+x^a)^2} - e^{-x} \right) \log{x} \, dx $$

Of course, we recognise the second term as a familiar definition of/easy-to-derive formula for $\gamma$. The first term we need to show is zero. But $$ \int \frac{ax^a\log{x}}{x(1+x^a)^2} \, dx = \frac{x^a\log{x}}{1+x^a} - \frac{1}{a}\log{(1+x^a)}, $$ which is continuous and tends to zero at both endpoints since $a>0$.

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The really interesting thing about this result in my opinion is that it shows the first term is a complete red herring: let $F$ be continuous and continuously differentiable on $(0,\infty)$ with the following properties:

• $ F(x) = 1 + o(1/\log{x})$ as $x \downarrow 0 $,


• $F(x) = o(x^{-\epsilon})$ as $x \uparrow \infty$ for some $\epsilon>0$,


• $\int_0^{\infty} F'(x) \log{x} \, dx = 0$

Then $$ \gamma = \int_0^{\infty} \left( F(x) - e^{-x} \right) \frac{dx}{x}. $$ The proof is essentially identical to the above: $$ \int_0^{\infty} ( F(x) - e^{-x} ) \frac{dx}{x} = [(F(x) - e^{-x}) \log{x}]_0^{\infty} - \int_0^{\infty} ( F'(x) + e^{-x} ) \log{x} \, dx = \int_0^{\infty} e^{-x} \log{x} \, dx, $$ The integral on the left exists by the first two conditions on $F$, which are also enough to ensure the boundary terms from the integration by parts go to zero.