Question Number 196852 by SANOGO last updated on 01/Sep/23
Answered by Mathspace last updated on 02/Sep/23
![∫_0 ^1 (t^(a−1) /(1+t^b ))dt =∫_0 ^1 t^(a−1) Σ_(n=0) ^∞ (−1)^n t^(nb) dt =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 t^(a+nb−1) dt =Σ_(n=0) ^∞ (−1)^n [(1/(a+nb))t^(a+nb) ]_0 ^1 =Σ_(n=0) ^∞ (((−1)^n )/(a+nb)) =Σ_(n=0) ^∞ (1/(a+2nb))−Σ_(n=0) ^∞ (1/(a+(2n+1)b)) =Σ_(n=0) ^∞ ((1/(a+2nb))−(1/(a+(2n+1)b))) =Σ_(n=0) ^∞ ((a+2nb+b−a−2nb)/((a+2nb)(a+(2n+1)b))) =bΣ_(n=0) ^∞ (1/(2b(n+(a/(2b)))2b((a/(2b))+(2n+1)(b/(2b))))) =(1/(4b))Σ_(n=0) ^∞ (1/((n+(a/(2b)))(n+(a/(2b))+(1/2)))) on rappelle que Σ_(n=0) ^∞ (1/((n+α)(n+β)))=((Ψ(α)−Φ(β))/(α−β)) (pour α≠β) S=(1/(4b))×((Ψ((a/(2b))+(1/2))−Ψ((a/(2b))))/((a/(2b))+(1/2)−(a/(2b)))) =(1/(2b)){Ψ((a/(2b))+(1/2))−Ψ((a/(2b)))} exemple a=b=1 ⇒∫_0 ^1 (dt/(1+t))=(1/2){Ψ(1)−Ψ((1/2))} =(1/2){−γ−(−γ−2ln2)} =(1/2)(2ln2)=ln2](https://www.tinkutara.com/question/Q196874.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}^{{b}} }{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{a}−\mathrm{1}} \sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {t}^{{nb}} {dt} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{a}+{nb}−\mathrm{1}} {dt} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \left[\frac{\mathrm{1}}{{a}+{nb}}{t}^{{a}+{nb}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{a}+{nb}} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{a}+\mathrm{2}{nb}}−\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{a}+\left(\mathrm{2}{n}+\mathrm{1}\right){b}} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{{a}+\mathrm{2}{nb}}−\frac{\mathrm{1}}{{a}+\left(\mathrm{2}{n}+\mathrm{1}\right){b}}\right) \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{{a}+\mathrm{2}{nb}+{b}−{a}−\mathrm{2}{nb}}{\left({a}+\mathrm{2}{nb}\right)\left({a}+\left(\mathrm{2}{n}+\mathrm{1}\right){b}\right)} \\ $$$$={b}\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{2}{b}\left({n}+\frac{{a}}{\mathrm{2}{b}}\right)\mathrm{2}{b}\left(\frac{{a}}{\mathrm{2}{b}}+\left(\mathrm{2}{n}+\mathrm{1}\right)\frac{{b}}{\mathrm{2}{b}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{b}}\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({n}+\frac{{a}}{\mathrm{2}{b}}\right)\left({n}+\frac{{a}}{\mathrm{2}{b}}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$${on}\:{rappelle}\:{que} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({n}+\alpha\right)\left({n}+\beta\right)}=\frac{\Psi\left(\alpha\right)−\Phi\left(\beta\right)}{\alpha−\beta} \\ $$$$\left({pour}\:\alpha\neq\beta\right) \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{4}{b}}×\frac{\Psi\left(\frac{{a}}{\mathrm{2}{b}}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\Psi\left(\frac{{a}}{\mathrm{2}{b}}\right)}{\frac{{a}}{\mathrm{2}{b}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{{a}}{\mathrm{2}{b}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{b}}\left\{\Psi\left(\frac{{a}}{\mathrm{2}{b}}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\Psi\left(\frac{{a}}{\mathrm{2}{b}}\right)\right\} \\ $$$${exemple}\:\:{a}={b}=\mathrm{1} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\mathrm{1}+{t}}=\frac{\mathrm{1}}{\mathrm{2}}\left\{\Psi\left(\mathrm{1}\right)−\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{−\gamma−\left(−\gamma−\mathrm{2}{ln}\mathrm{2}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{ln}\mathrm{2}\right)={ln}\mathrm{2} \\ $$
Commented by Mathspace last updated on 02/Sep/23
$$\Psi\:{est}\:{la}\:{fonction}\:{digamma} \\ $$
Commented by SANOGO last updated on 02/Sep/23
$${merci}\:{bien} \\ $$