Question Number 65383 by mathmax by abdo last updated on 29/Jul/19
$$\left.\mathrm{1}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\mathrm{1}+{e}^{{nx}} }\:\:\:{with}\:{n}\:{integr}\:{natural}\:\:{and}\:{n}\geqslant\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{conclude}\:{the}\:{value}\:{of}\:\sum_{{k}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}} \\ $$
Commented by mathmax by abdo last updated on 30/Jul/19
![1) let A_n =∫_0 ^∞ (dx/(1+e^(nx) )) changement e^(nx) =t vive nx =lnt ⇒ A_n =∫_1 ^(+∞) (dt/(nt(1+t))) =(1/n)∫_1 ^(+∞) {(1/t)−(1/(t+1))}dt =(1/n)[ln∣(t/(t+1))∣]_1 ^(+∞) =(1/n){−ln((1/2))} =((ln(2))/n) 2) we have A_n =∫_0 ^∞ (dx/(1+e^(nx) )) =∫_0 ^∞ (e^(−nx) /(1+e^(−nx) ))dx =∫_0 ^∞ e^(−nx) {Σ_(k=0) ^∞ (−1)^k e^(−knx) } =Σ_(k=0) ^∞ (−1)^k ∫_0 ^∞ e^(−(n+kn)x) dx =Σ_(k=0) ^∞ (−1)^k [−(1/(n(k+1)))e^(−(n+kn)x) ]_0 ^(+∞) =(1/n)Σ_(k=0) ^∞ (((−1)^k )/(k+1)) =(1/n) Σ_(k=1) ^∞ (((−1)^(k−1) )/k) but A_n =((ln(2))/n) ⇒Σ_(k=1) ^∞ (((−1)^(k−1) )/k) =ln(2).](https://www.tinkutara.com/question/Q65467.png)
$$\left.\mathrm{1}\right)\:{let}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\mathrm{1}+{e}^{{nx}} }\:\:{changement}\:\:{e}^{{nx}} \:={t}\:{vive}\:{nx}\:={lnt}\:\Rightarrow \\ $$$${A}_{{n}} =\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{dt}}{{nt}\left(\mathrm{1}+{t}\right)}\:=\frac{\mathrm{1}}{{n}}\int_{\mathrm{1}} ^{+\infty} \left\{\frac{\mathrm{1}}{{t}}−\frac{\mathrm{1}}{{t}+\mathrm{1}}\right\}{dt}\:=\frac{\mathrm{1}}{{n}}\left[{ln}\mid\frac{{t}}{{t}+\mathrm{1}}\mid\right]_{\mathrm{1}} ^{+\infty} \\ $$$$=\frac{\mathrm{1}}{{n}}\left\{−{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right\}\:=\frac{{ln}\left(\mathrm{2}\right)}{{n}} \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\mathrm{1}+{e}^{{nx}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{nx}} }{\mathrm{1}+{e}^{−{nx}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{nx}} \left\{\sum_{{k}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{k}} \:{e}^{−{knx}} \right\}\:=\sum_{{k}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{k}} \:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left({n}+{kn}\right){x}} {dx} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{k}} \left[−\frac{\mathrm{1}}{{n}\left({k}+\mathrm{1}\right)}{e}^{−\left({n}+{kn}\right){x}} \right]_{\mathrm{0}} ^{+\infty} \:=\frac{\mathrm{1}}{{n}}\sum_{{k}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{{n}}\:\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}}\:\:\:{but}\:{A}_{{n}} =\frac{{ln}\left(\mathrm{2}\right)}{{n}}\:\Rightarrow\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}}\:={ln}\left(\mathrm{2}\right). \\ $$