Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 68149 by ~ À ® @ 237 ~ last updated on 06/Sep/19

Explicit f(a)=Σ_(n=1) ^∞ (((−1)^n )/(n(an+1)))

$$\:{Explicit}\:\:\:{f}\left({a}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\left({an}+\mathrm{1}\right)}\:\:\:\: \\ $$

Commented by turbo msup by abdo last updated on 06/Sep/19

if a=0 f(0)=Σ_(n=1) ^∞ (((−1)^n )/n) =−ln(2) if a≠0 we have ((f(a))/a)=Σ_(n=1) ^∞ (((−1)^n )/(an(an+1))) =Σ_(n=1) ^∞ (−1)^n {(1/(an))−(1/(an+1))} =(1/a)Σ_(n=1) ^∞ (((−1)^n )/n)−Σ_(n=1) ^∞ (((−1)^n )/(an+1)) =−(1/a)ln(2)−Σ_(n=1) ^∞ (((−1)^n )/(an+1)) let s(x) =Σ_(n=1) ^∞ (((−1)^n )/(an+1))x^(an+1) with∣x∣<1 s(1)=Σ_(n=1) ^∞ (((−1)^n )/(an+1)) s^′ (x) =Σ_(n=1) ^∞ (−1)^n x^(an) =Σ_(n=1) ^∞ (−x^a )^n =(1/(1+x^a )) ⇒ s(x) =∫_0 ^x (dt/(1+t^a )) +c s(0)=0=c ⇒s(x)=∫_0 ^x (dt/(1+t^a )) ⇒ s(1) =∫_0 ^1 (dt/(1+t^a )) ⇒ ((f(a))/a) =−(1/a)ln(2)−∫_0 ^1 (dt/(1+t^a )) ⇒ f(a) =−ln(2)−a∫_0 ^1 (dt/(1+t^a )) be continued....

$${if}\:{a}=\mathrm{0}\:\:\:{f}\left(\mathrm{0}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}} \\ $$$$=−{ln}\left(\mathrm{2}\right) \\ $$$${if}\:{a}\neq\mathrm{0}\:\:\:{we}\:{have}\:\frac{{f}\left({a}\right)}{{a}}=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{an}\left({an}+\mathrm{1}\right)} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \left\{\frac{\mathrm{1}}{{an}}−\frac{\mathrm{1}}{{an}+\mathrm{1}}\right\} \\ $$$$=\frac{\mathrm{1}}{{a}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{an}+\mathrm{1}} \\ $$$$=−\frac{\mathrm{1}}{{a}}{ln}\left(\mathrm{2}\right)−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{an}+\mathrm{1}} \\ $$$${let}\:{s}\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{an}+\mathrm{1}}{x}^{{an}+\mathrm{1}} \:\:{with}\mid{x}\mid<\mathrm{1} \\ $$$${s}\left(\mathrm{1}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{an}+\mathrm{1}} \\ $$$${s}^{'} \left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\left(−\mathrm{1}\right)^{{n}} {x}^{{an}} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \left(−{x}^{{a}} \right)^{{n}} \:\:=\frac{\mathrm{1}}{\mathrm{1}+{x}^{{a}} }\:\Rightarrow \\ $$$${s}\left({x}\right)\:=\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{{dt}}{\mathrm{1}+{t}^{{a}} }\:+{c} \\ $$$${s}\left(\mathrm{0}\right)=\mathrm{0}={c}\:\Rightarrow{s}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \:\frac{{dt}}{\mathrm{1}+{t}^{{a}} }\:\Rightarrow \\ $$$${s}\left(\mathrm{1}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{1}+{t}^{{a}} }\:\:\Rightarrow \\ $$$$\frac{{f}\left({a}\right)}{{a}}\:=−\frac{\mathrm{1}}{{a}}{ln}\left(\mathrm{2}\right)−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{\mathrm{1}+{t}^{{a}} }\:\Rightarrow \\ $$$${f}\left({a}\right)\:=−{ln}\left(\mathrm{2}\right)−{a}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{\mathrm{1}+{t}^{{a}} } \\ $$$${be}\:{continued}.... \\ $$

Commented by mathmax by abdo last updated on 06/Sep/19

error from line 10 s^′ (x) =Σ_(n=1) ^∞ (−x^a )^n =−x^a Σ_(n=1) ^∞ (−x^a )^(n−1) =−x^a Σ_(n=0) ^∞ (−x^a )^n =((−x^a )/(1+x^a )) ⇒s(x) =−∫_0 ^x (t^a /(1+t^a ))dt +c s(0) =0 =c ⇒s(x) =−∫_0 ^x (t^a /(1+t^a ))dt ⇒ Σ_(n=1) ^∞ (((−1)^n )/(an+1)) =s(1) =−∫_0 ^1 (t^a /(1+t^a ))dt =−∫_0 ^1 ((1+t^a −1)/(1+t^a ))dt =−1 +∫_0 ^1 (dt/(1+t^a )) be continued...

$${error}\:{from}\:{line}\:\mathrm{10}\:\:{s}^{'} \left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \left(−{x}^{{a}} \right)^{{n}} \:=−{x}^{{a}} \sum_{{n}=\mathrm{1}} ^{\infty} \left(−{x}^{{a}} \right)^{{n}−\mathrm{1}} \\ $$$$=−{x}^{{a}} \sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−{x}^{{a}} \right)^{{n}} \:=\frac{−{x}^{{a}} }{\mathrm{1}+{x}^{{a}} }\:\Rightarrow{s}\left({x}\right)\:=−\int_{\mathrm{0}} ^{{x}} \:\frac{{t}^{{a}} }{\mathrm{1}+{t}^{{a}} }{dt}\:+{c} \\ $$$${s}\left(\mathrm{0}\right)\:=\mathrm{0}\:={c}\:\Rightarrow{s}\left({x}\right)\:=−\int_{\mathrm{0}} ^{{x}} \:\frac{{t}^{{a}} }{\mathrm{1}+{t}^{{a}} }{dt}\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{an}+\mathrm{1}}\:={s}\left(\mathrm{1}\right)\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{t}^{{a}} }{\mathrm{1}+{t}^{{a}} }{dt}\:\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}+{t}^{{a}} −\mathrm{1}}{\mathrm{1}+{t}^{{a}} }{dt} \\ $$$$=−\mathrm{1}\:+\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{\mathrm{1}+{t}^{{a}} }\:\:\:\:{be}\:{continued}... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com