TAOCP 5.2.2 Exercise 43
We prove the identity \[ \int_0^1 \frac{e^{-y}-1}{y}\,dy \;+\; \int_1^\infty \frac{e^{-y}}{y}\,dy \;=\; -\gamma, \] where \(\gamma\) is the Euler-Mascheroni constant.
Section 5.2.2: Sorting by Exchanging
Exercise 43. [HM21] Prove that $\int_0^y y^{-1}(e^{-y} - 1),dy + \int_1^\infty y^{-1} e^{-y},dy = -\gamma$. [Hint: Consider $\lim_{a \to 0} y^{a-1}$.]
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Solution
We prove the identity [ \int_0^1 \frac{e^{-y}-1}{y},dy ;+; \int_1^\infty \frac{e^{-y}}{y},dy ;=; -\gamma, ] where (\gamma) is the Euler-Mascheroni constant. (The upper limit of the first integral is (1); the notation (\int_0^y) in the exercise statement is a typographical error.)
For (a>0) define [ I(a) = \int_0^1 (e^{-y}-1),y^{a-1},dy ;+; \int_1^\infty e^{-y},y^{a-1},dy. ] Both integrals converge absolutely for (a>0). As (a\to 0^+), the integrands converge pointwise to the integrands of the desired integrals, and they are dominated by integrable functions: \begin{itemize} \item On ([0,1]), (|(e^{-y}-1)y^{a-1}| \le C) for some constant (C) because (e^{-y}-1 \sim -y). \item On ([1,\infty)), (e^{-y}y^{a-1} \le e^{-y}/y) for (a\ge 0), and (\int_1^\infty e^{-y}/y,dy) converges. \end{itemize} Hence by the dominated convergence theorem, [ \lim_{a\to 0^+} I(a) = \int_0^1 \frac{e^{-y}-1}{y},dy ;+; \int_1^\infty \frac{e^{-y}}{y},dy. ]
Now express (I(a)) in terms of the Gamma function (\Gamma(a)=\int_0^\infty e^{-y}y^{a-1},dy): [ \int_0^1 e^{-y}y^{a-1},dy = \Gamma(a) - \int_1^\infty e^{-y}y^{a-1},dy. ] Therefore [ I(a) = \left(\Gamma(a) - \int_1^\infty e^{-y}y^{a-1},dy\right) - \int_0^1 y^{a-1},dy + \int_1^\infty e^{-y}y^{a-1},dy = \Gamma(a) - \frac{1}{a}. ]
The Laurent expansion of (\Gamma(a)) at (a=0) is [ \Gamma(a) = \frac{1}{a} - \gamma + O(a) \qquad (a\to 0). ] Thus [ I(a) = -\gamma + O(a), ] and taking the limit (a\to 0^+) yields [ \int_0^1 \frac{e^{-y}-1}{y},dy ;+; \int_1^\infty \frac{e^{-y}}{y},dy = -\gamma. ]
This completes the proof. ∎