Question Number 46610 by maxmathsup by imad last updated on 29/Oct/18
$${let}\:{f}_{{n}} \left({x}\right)={e}^{−{nx}} −\mathrm{2}{e}^{−\mathrm{2}{nx}} \:\:{with}\:{x}\:{from}\left[\mathrm{0},+\infty\left[\right.\right. \\ $$$$\left.\mathrm{1}\right){calculate}\:\int_{\mathrm{0}} ^{\infty} \:{f}_{{n}} \left({x}\right){dx}\:\:{and}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(\int_{\mathrm{0}} ^{\infty} \:{f}_{{n}} \left({x}\right){dx}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{S}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:{f}_{{n}} \left({x}\right)\:\:{and}\:\int_{\mathrm{0}} ^{\infty} \:{S}\left({x}\right){dx} \\ $$
Commented by maxmathsup by imad last updated on 05/Nov/18
![1)∫_0 ^∞ f_n (x)dx =∫_0 ^∞ (e^(−nx) −2e^(−2nx) )dx =∫_0 ^∞ e^(−nx) dx−2 ∫_0 ^∞ e^(−2nx) dx =[−(1/n) e^(−nx) ]_0 ^∞ −2 [−(1/(2n)) e^(−2nx) ]_0 ^∞ =(1/n) −2((1/(2n))) =0 ⇒Σ_(n=0) ^∞ (∫_0 ^∞ f_n (x)dx)=0](https://www.tinkutara.com/question/Q47165.png)
$$\left.\mathrm{1}\right)\int_{\mathrm{0}} ^{\infty} \:\:{f}_{{n}} \left({x}\right){dx}\:=\int_{\mathrm{0}} ^{\infty} \:\left({e}^{−{nx}} −\mathrm{2}{e}^{−\mathrm{2}{nx}} \right){dx}\:=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{nx}} {dx}−\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\mathrm{2}{nx}} {dx} \\ $$$$=\left[−\frac{\mathrm{1}}{{n}}\:{e}^{−{nx}} \right]_{\mathrm{0}} ^{\infty} \:−\mathrm{2}\:\left[−\frac{\mathrm{1}}{\mathrm{2}{n}}\:{e}^{−\mathrm{2}{nx}} \right]_{\mathrm{0}} ^{\infty} \:=\frac{\mathrm{1}}{{n}}\:−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}{n}}\right)\:=\mathrm{0}\:\:\Rightarrow\sum_{{n}=\mathrm{0}} ^{\infty} \left(\int_{\mathrm{0}} ^{\infty} {f}_{{n}} \left({x}\right){dx}\right)=\mathrm{0} \\ $$
Commented by maxmathsup by imad last updated on 05/Nov/18
![2) we have S(x)=Σ_(n=0) ^∞ e^(−nx) −2Σ_(n=0) ^∞ e^(−2nx) =Σ_(n=0) ^∞ (e^(−x) )^n −2 Σ_(n=0) ^(∞ ) (e^(−2x) )^n and for x>0 we get S(x)=(1/(1−e^(−x) )) −2 (1/(1−e^(−2x) )) ∫ S(x)dx =∫ (dx/(1−e^(−x) )) −2 ∫ (dx/(1−e^(−2x) )) +c but ∫ (dx/(1−e^(−x) )) =∫ (e^x /(e^x −1)) dx =ln∣e^x −1∣ ∫ ((2dx)/(1−e^(−2x) )) =∫ ((2e^(2x) )/(e^(2x) −1)) dx =ln∣e^(2x) −1∣ ⇒∫ S(x)dx=ln∣((e^x −1)/(e^(2x) −1))∣ +c =ln∣(1/(e^x +1))∣ +c ⇒∫_0 ^∞ S(x)dx =[−ln(e^x +1)]_0 ^(+∞) =−∞ .](https://www.tinkutara.com/question/Q47167.png)
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{S}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{nx}} −\mathrm{2}\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−\mathrm{2}{nx}} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left({e}^{−{x}} \right)^{{n}} \:−\mathrm{2}\:\sum_{{n}=\mathrm{0}} ^{\infty\:} \:\left({e}^{−\mathrm{2}{x}} \right)^{{n}} \:\:\:{and}\:{for}\:{x}>\mathrm{0}\:{we}\:{get} \\ $$$${S}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{x}} }\:−\mathrm{2}\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{2}{x}} } \\ $$$$\int\:{S}\left({x}\right){dx}\:=\int\:\:\frac{{dx}}{\mathrm{1}−{e}^{−{x}} }\:−\mathrm{2}\:\int\:\:\frac{{dx}}{\mathrm{1}−{e}^{−\mathrm{2}{x}} }\:+{c}\:\:{but} \\ $$$$\int\:\:\frac{{dx}}{\mathrm{1}−{e}^{−{x}} }\:=\int\:\:\frac{{e}^{{x}} }{{e}^{{x}} −\mathrm{1}}\:{dx}\:={ln}\mid{e}^{{x}} −\mathrm{1}\mid \\ $$$$\int\:\:\:\frac{\mathrm{2}{dx}}{\mathrm{1}−{e}^{−\mathrm{2}{x}} }\:=\int\:\frac{\mathrm{2}{e}^{\mathrm{2}{x}} }{{e}^{\mathrm{2}{x}} −\mathrm{1}}\:{dx}\:={ln}\mid{e}^{\mathrm{2}{x}} −\mathrm{1}\mid\:\Rightarrow\int\:{S}\left({x}\right){dx}={ln}\mid\frac{{e}^{{x}} −\mathrm{1}}{{e}^{\mathrm{2}{x}} −\mathrm{1}}\mid\:+{c} \\ $$$$={ln}\mid\frac{\mathrm{1}}{{e}^{{x}} \:+\mathrm{1}}\mid\:+{c}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:{S}\left({x}\right){dx}\:=\left[−{ln}\left({e}^{{x}} \:+\mathrm{1}\right)\right]_{\mathrm{0}} ^{+\infty} \:=−\infty\:. \\ $$$$ \\ $$