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

$$letgivex>0$$ $$1)provethat{\int }_0^1\frac{dt}{1+t^x}=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{nx+1}$$ $$2)find\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n+1}and\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{2n+1}$$ $$3)find\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{3n+1}.$$

Commented byabdo imad last updated on 16/Feb/18

$$1)fort\in ]0,1]t^x=e^{xlnt}<1\Rightarrow {\int }_0^1\frac{dt}{1+t^x}={\int }_0^1\left(\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{t}^{nx}\right)dt$$ $$=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{\int }_0^1{t}^{nx}dt=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{nx+1}$$ $$2)wehaveprovedthatA\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{nx+1}={\int }_0^1\frac{dt}{1+t^x}\Rightarrow$$ $$\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n+1}=A\left(1\right)={\int }_0^1\frac{dt}{1+t}={\left[ln(1+t)\right]}_0^1=ln\left(2\right)$$ $$\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{2n+1}=A\left(2\right)={\int }_0^1\frac{dt}{1+t^2}={\left[arctant\right]}_0^1=\frac{\pi }{4}$$ $$3)wehave\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{3n+1}=A\left(3\right)={\int }_0^1\frac{dt}{1+t^3}$$ $$wehave{\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^3}={\int }_0^1\frac{dt}{1+t^3}+{\int }_1^{+\mathrm{\infty }}\frac{dt}{1+t^3}thech.t=\frac{1}{u}give$$ $${\int }_1^{\mathrm{\infty }}\frac{dt}{1+t^3}={\int }_0^1\frac{1}{1+\frac{1}{u^3}}\frac{du}{u^2}={\int }_0^1\frac{du}{u^2+\frac{1}{u}}={\int }_0^1\frac{udu}{1+u^2}$$ $$=\frac{1}{2}{\left[ln(1+u^2)\right]}_0^1=\frac{1}{2}ln2thech.t^3=ugive$$ $${\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^3}={\int }_0^{\mathrm{\infty }}\frac{1}{1+u}\frac{1}{3}{u}^{\frac{1}{3}-1}du=\frac{1}{3}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{1}{3}-1}}{1+u}du$$ $$=\frac{1}{3}\frac{\pi }{sin\left(\frac{\pi }{3}\right)}=\frac{\pi }{3}\frac{1}{\frac{\sqrt{3}}{2}}=\frac{2\pi }{3\sqrt{3}}\Rightarrow$$ $${\int }_0^1\frac{dt}{1+t^3}={\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^3}-{\int }_1^{+\mathrm{\infty }}\frac{dt}{1+t^3}=\frac{2\pi }{3\sqrt{3}}-\frac{1}{2}ln\left(2\right)\Rightarrow$$ $$\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{3n+1}=\frac{2\pi }{3\sqrt{3}}-\frac{1}{2}ln\left(2\right).$$

Commented byabdo imad last updated on 16/Feb/18

$$forQ3)wehaveusedtheresult{\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{1+t}dt=\frac{\pi }{sin\left(\pi a\right)}$$ $$with0<a<1.$$

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