Question Number 29978 by abdo imad last updated on 14/Feb/18
$${let}\:{give}\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{\mathrm{1}+{t}^{{x}} }=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{nx}+\mathrm{1}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:\:{and}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\left.\mathrm{3}\right)\:{find}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}{n}+\mathrm{1}}\:. \\ $$
Commented by abdo 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).](https://www.tinkutara.com/question/Q30074.png)
$$\left.\mathrm{1}\left.\right)\left.\:{for}\:{t}\in\right]\mathrm{0},\mathrm{1}\right]\:\:{t}^{{x}} \:={e}^{{xlnt}} <\mathrm{1}\:\Rightarrow\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{1}+{t}^{{x}} }=\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {t}^{{nx}} \right){dt} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{{nx}} {dt}=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{nx}\:+\mathrm{1}} \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{proved}\:{that}\:{A}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{nx}+\mathrm{1}}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\mathrm{1}+{t}^{{x}} }\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}={A}\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{\mathrm{1}+{t}}=\left[{ln}\left(\mathrm{1}+{t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:\:={ln}\left(\mathrm{2}\right) \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}={A}\left(\mathrm{2}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }=\left[{arctant}\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\pi}{\mathrm{4}} \\ $$$$\left.\mathrm{3}\right){we}\:{have}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}{n}+\mathrm{1}}=\:{A}\left(\mathrm{3}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{3}} }\: \\ $$$${we}\:{have}\:\:\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{3}} }\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{3}} }\:+\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{3}} }\:{the}\:{ch}.\:{t}=\frac{\mathrm{1}}{{u}}\:{give} \\ $$$$\int_{\mathrm{1}} ^{\infty} \:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{3}} }=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{u}^{\mathrm{3}} }}\:\frac{{du}}{{u}^{\mathrm{2}} }=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{du}}{{u}^{\mathrm{2}} +\frac{\mathrm{1}}{{u}}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{udu}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2}\:\:\:{the}\:{ch}.\:{t}^{\mathrm{3}} ={u}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{3}} }\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{1}+{u}}\frac{\mathrm{1}}{\mathrm{3}}{u}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} {du}\:=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \:\frac{{u}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} }{\mathrm{1}+{u}}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\:\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{3}}\right)}=\frac{\pi}{\mathrm{3}}\:\frac{\mathrm{1}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:=\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{3}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{3}} }\:−\:\int_{\mathrm{1}} ^{+\infty} \:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{3}} }=\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}{n}+\mathrm{1}}=\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right). \\ $$
Commented by abdo imad last updated on 16/Feb/18
$$\left.{for}\:{Q}\:\mathrm{3}\right)\:{we}\:{have}\:{used}\:{the}\:{result}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}=\frac{\pi}{{sin}\left(\pi{a}\right)} \\ $$$${with}\:\mathrm{0}<{a}<\mathrm{1}\:. \\ $$