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Question Number 29978 by abdo imad last updated on 14/Feb/18

let give x>0  1) prove that   ∫_0 ^1     (dt/(1+t^x ))= Σ_(n=0) ^∞    (((−1)^n )/(nx+1))  2) find Σ_(n=0) ^∞   (((−1)^n )/(n+1))  and Σ_(n=0) ^∞   (((−1)^n )/(2n+1))  3) find Σ_(n=1) ^∞    (((−1)^n )/(3n+1)) .

$${let}\:{give}\:{x}>\mathrm{0} \\ $$ $$\left.\mathrm{1}\right)\:{prove}\:{that}\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{\mathrm{1}+{t}^{{x}} }=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{nx}+\mathrm{1}} \\ $$ $$\left.\mathrm{2}\right)\:{find}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:\:{and}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$ $$\left.\mathrm{3}\right)\:{find}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}{n}+\mathrm{1}}\:. \\ $$

Commented byabdo imad last updated on 16/Feb/18

1) for t∈]0,1]  t^x  =e^(xlnt) <1 ⇒ ∫_0 ^1   (dt/(1+t^x ))=∫_0 ^1  (Σ_(n=0) ^∞ (−1)^n t^(nx) )dt  =Σ_(n=0) ^∞  (−1)^n  ∫_0 ^1  t^(nx) dt=Σ_(n=0) ^∞  (((−1)^n )/(nx +1))  2) we have proved that A(x)=Σ_(n=0) ^∞  (((−1)^n )/(nx+1))=∫_0 ^1    (dt/(1+t^x ))⇒  Σ_(n=0) ^∞   (((−1)^n )/(n+1))=A(1)=∫_0 ^1  (dt/(1+t))=[ln(1+t)]_0 ^1   =ln(2)  Σ_(n=0) ^∞   (((−1)^n )/(2n+1))=A(2)=∫_0 ^1   (dt/(1+t^2 ))=[arctant]_0 ^1 =(π/4)  3)we have Σ_(n=0) ^∞  (((−1)^n )/(3n+1))= A(3) = ∫_0 ^1   (dt/(1+t^3 ))   we have  ∫_0 ^∞ (dt/(1+t^3 )) = ∫_0 ^1   (dt/(1+t^3 )) +∫_1 ^(+∞)   (dt/(1+t^3 )) the ch. t=(1/u) give  ∫_1 ^∞   (dt/(1+t^3 ))=∫_0 ^1    (1/(1+(1/u^3 ))) (du/u^2 )= ∫_0 ^1      (du/(u^2 +(1/u))) =∫_0 ^1  ((udu)/(1+u^2 ))  =(1/2)[ln(1+u^2 )]_0 ^1  =(1/2)ln2   the ch. t^3 =u give  ∫_0 ^∞  (dt/(1+t^3 )) = ∫_0 ^∞   (1/(1+u))(1/3)u^((1/3)−1) du =(1/3)∫_0 ^∞  (u^((1/3)−1) /(1+u))du  =(1/3) (π/(sin((π/3))))=(π/3) (1/((√3)/2)) =((2π)/(3(√3))) ⇒  ∫_0 ^1   (dt/(1+t^3 )) =∫_0 ^∞   (dt/(1+t^3 )) − ∫_1 ^(+∞)  (dt/(1+t^3 ))=((2π)/(3(√3))) −(1/2)ln(2)⇒  Σ_(n=0) ^∞   (((−1)^n )/(3n+1))=((2π)/(3(√3))) −(1/2)ln(2).

$$\left.\mathrm{1}\left.\right)\left.\:{for}\:{t}\in\right]\mathrm{0},\mathrm{1}\right]\:\:{t}^{{x}} \:={e}^{{xlnt}} <\mathrm{1}\:\Rightarrow\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{1}+{t}^{{x}} }=\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {t}^{{nx}} \right){dt} \\ $$ $$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{{nx}} {dt}=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{nx}\:+\mathrm{1}} \\ $$ $$\left.\mathrm{2}\right)\:{we}\:{have}\:{proved}\:{that}\:{A}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{nx}+\mathrm{1}}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\mathrm{1}+{t}^{{x}} }\Rightarrow \\ $$ $$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}={A}\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{\mathrm{1}+{t}}=\left[{ln}\left(\mathrm{1}+{t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:\:={ln}\left(\mathrm{2}\right) \\ $$ $$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}={A}\left(\mathrm{2}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }=\left[{arctant}\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\pi}{\mathrm{4}} \\ $$ $$\left.\mathrm{3}\right){we}\:{have}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}{n}+\mathrm{1}}=\:{A}\left(\mathrm{3}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{3}} }\: \\ $$ $${we}\:{have}\:\:\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{3}} }\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{3}} }\:+\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{3}} }\:{the}\:{ch}.\:{t}=\frac{\mathrm{1}}{{u}}\:{give} \\ $$ $$\int_{\mathrm{1}} ^{\infty} \:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{3}} }=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{u}^{\mathrm{3}} }}\:\frac{{du}}{{u}^{\mathrm{2}} }=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{du}}{{u}^{\mathrm{2}} +\frac{\mathrm{1}}{{u}}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{udu}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2}\:\:\:{the}\:{ch}.\:{t}^{\mathrm{3}} ={u}\:{give} \\ $$ $$\int_{\mathrm{0}} ^{\infty} \:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{3}} }\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{1}+{u}}\frac{\mathrm{1}}{\mathrm{3}}{u}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} {du}\:=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \:\frac{{u}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} }{\mathrm{1}+{u}}{du} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{3}}\:\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{3}}\right)}=\frac{\pi}{\mathrm{3}}\:\frac{\mathrm{1}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:=\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$ $$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{3}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{3}} }\:−\:\int_{\mathrm{1}} ^{+\infty} \:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{3}} }=\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\Rightarrow \\ $$ $$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}{n}+\mathrm{1}}=\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right). \\ $$

Commented byabdo imad last updated on 16/Feb/18

for Q 3) we have used the result ∫_0 ^∞  (t^(a−1) /(1+t))dt=(π/(sin(πa)))  with 0<a<1 .

$$\left.{for}\:{Q}\:\mathrm{3}\right)\:{we}\:{have}\:{used}\:{the}\:{result}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}=\frac{\pi}{{sin}\left(\pi{a}\right)} \\ $$ $${with}\:\mathrm{0}<{a}<\mathrm{1}\:. \\ $$

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