Ivan Shishkin, Rye (1878)

Problems/Riemann integration

une intégrale très très très sympa

by mathman·
50
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γ\gamma désigne la constante d’Euler ≈0,577215…\approx0{,}577215\ldots qu’on peut définir par γ=lim⁡n→+∞(∑k=1n1k−ln⁡(n))\gamma=\displaystyle\lim_{n \to +\infty}(\sum_{k=1}^{n}\frac{1}{k}- \ln(n))
Montrez que :
∫011−e−t−e−1/tt dt=γ.\displaystyle\int_{0}^{1}\frac{1-\mathrm{e}^{-t}-\mathrm{e}^{-1/t}}{t}\,\mathrm{d}t=\gamma.

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Solution by mathman

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tout d’abord, on réécrit:

∫011−e−t−e−1/tt dt\displaystyle\int_{0}^{1}\frac{1-\mathrm{e}^{-t}-\mathrm{e}^{-1/t}}{t}\,\mathrm{d}t

sous la forme :
∫011−e−tt dt−∫1+∞e−tt dt\displaystyle\int_{0}^{1}\frac{1-\mathrm{e}^{-t}}{t}\,\mathrm{d}t-\displaystyle\int_{1}^{+\infty}\frac{\mathrm{e}^{-t}}{t}\,\mathrm{d}t

ensuite on utilise la série n−n22.2!+n33.3!−n44.4!⋯n-\frac{n^{2}}{2.2!}+\frac{n^{3}}{3.3!}-\frac{n^{4}}{4.4!}\cdots pour écrire:

n−n22.2!+n33.3!−n44.4!⋯=∫0n1−e−tt dt=∫011−e−tt dt+ln⁡(n)−∫1ne−tt dtn-\frac{n^{2}}{2.2!}+\frac{n^{3}}{3.3!}-\frac{n^{4}}{4.4!}\cdots=\displaystyle\int_{0}^{n}\frac{1-\mathrm{e}^{-t}}{t}\,\mathrm{d}t=\displaystyle\int_{0}^{1}\frac{1-\mathrm{e}^{-t}}{t}\,\mathrm{d}t+\ln(n)-\int_{1}^{n}\frac{\mathrm{e}^{-t}}{t}\,\mathrm{d}t

n−n22.2!+n33.3!−n44.4!⋯=∫011−e−tt dt+ln⁡(n)−∫1+∞e−tt dt+∫n+∞e−tt dtn-\frac{n^{2}}{2.2!}+\frac{n^{3}}{3.3!}-\frac{n^{4}}{4.4!}\cdots=\displaystyle\int_{0}^{1}\frac{1-\mathrm{e}^{-t}}{t}\,\mathrm{d}t+\ln(n)-\int_{1}^{+\infty}\frac{\mathrm{e}^{-t}}{t}\,\mathrm{d}t+\int_{n}^{+\infty}\frac{\mathrm{e}^{-t}}{t}\,\mathrm{d}t

n−n22.2!+n33.3!−n44.4!⋯=∫011−e−tt dt+ln⁡(n)−∫1+∞e−tt dt+∫1+∞e−tnt dtn-\frac{n^{2}}{2.2!}+\frac{n^{3}}{3.3!}-\frac{n^{4}}{4.4!}\cdots=\displaystyle\int_{0}^{1}\frac{1-\mathrm{e}^{-t}}{t}\,\mathrm{d}t+\ln(n)-\int_{1}^{+\infty}\frac{\mathrm{e}^{-t}}{t}\,\mathrm{d}t+\int_{1}^{+\infty}\frac{\mathrm{e}^{-tn}}{t}\,\mathrm{d}t

on fait ainsi apparaitre, dans le second membre, l’expression recherchée:
∫011−e−tt dt−∫1+∞e−tt dt=∫011−e−t−e−1/tt dt\displaystyle\int_{0}^{1}\frac{1-\mathrm{e}^{-t}}{t}\,\mathrm{d}t-\displaystyle\int_{1}^{+\infty}\frac{\mathrm{e}^{-t}}{t}\,\mathrm{d}t=\displaystyle\int_{0}^{1}\frac{1-\mathrm{e}^{-t}-\mathrm{e}^{-1/t}}{t}\,\mathrm{d}t

ensuite on utilise la somme ∑k=1+∞(1k−1k+n)\displaystyle \sum_{k=1}^{+\infty}(\frac{1}{k}-\frac{1}{k+n}) pour écrire:

∑k=1+∞(1k−1k+n)=∫0+∞1−e−tnet−1 dt=∫011−tn1−t dt=∫011−(1−t)nt dt\displaystyle \sum_{k=1}^{+\infty}(\frac{1}{k}-\frac{1}{k+n})=\int_{0}^{+\infty}\frac{1-\mathrm{e}^{-tn}}{\mathrm{e}^{t}-1}\,\mathrm{d}t=\int_{0}^{1}\frac{1-\mathrm{t}^{n}}{1-t}\,\mathrm{d}t=\int_{0}^{1}\frac{1-(1-t)^{n}}{t}\,\mathrm{d}t

et comme:

n−n22.2!+n33.3!−n44.4!⋯=∫0n1−e−tt dt=∫011−e−tnt dtn-\frac{n^{2}}{2.2!}+\frac{n^{3}}{3.3!}-\frac{n^{4}}{4.4!}\cdots=\displaystyle\int_{0}^{n}\frac{1-\mathrm{e}^{-t}}{t}\,\mathrm{d}t=\displaystyle\int_{0}^{1}\frac{1-\mathrm{e}^{-tn}}{t}\,\mathrm{d}t

on en déduit que:

∑k=1+∞(1k−1k+n)=n−n22.2!+n33.3!−n44.4!⋯+∫01e−tn−(1−t)nt dt\displaystyle \sum_{k=1}^{+\infty}(\frac{1}{k}-\frac{1}{k+n})=n-\frac{n^{2}}{2.2!}+\frac{n^{3}}{3.3!}-\frac{n^{4}}{4.4!}\cdots+\int_{0}^{1}\frac{\mathrm{e}^{-tn}-(1-t)^{n}}{t}\,\mathrm{d}t

et finalement que:

∑k=1+∞(1k−1k+n)=ln⁡(n)+∫011−e−tt dt−∫1+∞e−tt dt+∫1+∞e−tnt dt+∫01e−tn−(1−t)nt dt\displaystyle \sum_{k=1}^{+\infty}(\frac{1}{k}-\frac{1}{k+n})=\displaystyle\ln(n)+\int_{0}^{1}\frac{1-\mathrm{e}^{-t}}{t}\,\mathrm{d}t-\int_{1}^{+\infty}\frac{\mathrm{e}^{-t}}{t}\,\mathrm{d}t+\int_{1}^{+\infty}\frac{\mathrm{e}^{-tn}}{t}\,\mathrm{d}t+\int_{0}^{1}\frac{\mathrm{e}^{-tn}-(1-t)^{n}}{t}\,\mathrm{d}t

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