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Question Number 115273 by mnjuly1970 last updated on 24/Sep/20

$$...advancedmathematics...$$$$$$$$evaluate:::$$$$$$$$\mathrm{\Delta }={\int }_0^{\mathrm{\infty }}\frac{cos\left(ln\left(x\right)\right)}{{(x+1)}^2}dx=???$$$$$$$$...m.n.july.1970...$$$$$$

Answered by Bird last updated on 24/Sep/20

$$A={\int }_0^{\mathrm{\infty }}\frac{cos\left(lnx\right)}{{(x+1)}^2}dx\Rightarrow$$$$A={\int }_0^1\frac{cos\left(lnx\right)}{{(x+1)}^2}dx+{\int }_1^{\mathrm{\infty }}\frac{cos\left(lnx\right)}{{(x+1)}^2}dx$$$$but{\int }_1^{\mathrm{\infty }}\frac{cos\left(lnx\right)}{{(x+1)}^2}dx{=}_{x=\frac{1}{t}}$$$$-{\int }_0^1\frac{cos\left(lnt\right)}{{(\frac{1}{t}+1)}^2}(-\frac{dt}{t^2})$$$$={\int }_0^1\frac{cos\left(lnx\right)}{{(x+1)}^2}\Rightarrow A=2{\int }_0^1\frac{cos\left(lnx\right)}{{(x+1)}^2}dx$$$$wehave\frac{1}{x+1}=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{x}^n$$$$\Rightarrow -\frac{1}{{(x+1)}^2}=\sum _{n=1}^{\mathrm{\infty }}{(-1)}^n{nx}^{n-1}$$$$\Rightarrow \frac{1}{{(x+1)}^2}=\sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n-1}{nx}^{n-1}$$$$=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n(n+1)x^n\Rightarrow$$$$A=2{\int }_0^1\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n(n+1)x^{n}cos\left(lnx\right)dx$$$$=2\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n(n+1){\int }_0^1{x}^{n}cos\left(lnx\right)dx$$$$U_n={\int }_0^1{x}^{n}cos\left(lnx\right)dx$$$${=}_{lnx=-t}-{\int }_0^{\mathrm{\infty }}{e}^{-nt}cos\left(t\right)(-e^{-t})dt$$$$={\int }_0^{\mathrm{\infty }}{e}^{-(n+1)t}cos\left(t\right)dt$$$$=Re\left({\int }_0^{\mathrm{\infty }}{e}^{-(n+1)t}{e}^{it}dt\right)but$$$${\int }_0^{\mathrm{\infty }}{e}^{(-(n+1)+i)t}dt$$$$={\left[\frac{1}{-(n+1)+i}{e}^{(-(n+1)+i)t}\right]}_0^{\mathrm{\infty }}$$$$=\frac{1}{n+1-i}=\frac{n+1+i}{{(n+1)}^2+1}\Rightarrow$$$$U_n=\frac{n+1}{{(n+1)}^2+1}\Rightarrow$$$$A=2\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n(n+1)\times \frac{n+1}{{(n+1)}^2+1}$$$$=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n\frac{{(n+1)}^2}{{(n+1)}^2+1}$$$$=\sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n-1}.\frac{n^2}{n^2+1}$$$$=\sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n-1}(1-\frac{1}{n^2+1})$$$$=\sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n-1}(\to diverges)+\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2+1}\left(converges\right)$$$$perhspsthisintegralisdivergent..!$$

Commented by mnjuly1970 last updated on 24/Sep/20

Commented by mnjuly1970 last updated on 24/Sep/20

$$integralisconvergent\Uparrow \Uparrow \Uparrow$$

Commented by Dwaipayan Shikari last updated on 24/Sep/20

$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}-1}=1-1+1-1+1-1+...$$$$\mathrm{Sum}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{exist}\mathrm{or}$$$$1-1+1-1+1-1+1-1=\frac{1}{2}\left(\mathrm{Analytical}\mathrm{continuation}\right)$$

Commented by Olaf last updated on 24/Sep/20

$${\mathrm{U}}_n={\int }_0^1{x}^n\cos \left(\ln x\right)dx={\int }_0^1{x}^{n+1}\frac{\cos \left(\ln x\right)}{x}dx$$$${\mathrm{U}}_n={\left[x^{n+1}\sin \left(\ln x\right)\right]}_0^1-{\int }_0^1(n+1)x^n\sin \left(\ln x\right)dx$$$${\mathrm{U}}_n=-(n+1){\int }_0^1{x}^{n+1}\frac{\sin \left(\ln x\right)}{x}dx$$$${\mathrm{U}}_n=-(n+1)\{{[-x^{n+1}\cos \left(\ln x\right)]}_0^1+(n+1){\int }_0^1{x}^n\cos \left(\ln x\right)dx\}$$$${\mathrm{U}}_n=-(n+1)(-1+(n+1){\mathrm{U}}_n)$$$$(1+{(n+1)}^2){\mathrm{U}}_n=n+1$$$${\mathrm{U}}_n=\frac{n+1}{{(n+1)}^2+1}$$$$maybe...$$$$$$

Commented by Bird last updated on 24/Sep/20

$$postyouanswersir$$

Commented by mnjuly1970 last updated on 24/Sep/20

$$ok.isentmysolution.$$

Commented by mnjuly1970 last updated on 24/Sep/20

Commented by Bird last updated on 25/Sep/20

$$okthanks$$

Answered by mnjuly1970 last updated on 24/Sep/20

$$solution$$$$\mathrm{\Delta }=\mathcal{R}e{\int }_0^{\mathrm{\infty }}\frac{e^{iln\left(x\right)}}{{(x+1)}^2}dx=\mathcal{R}e{\int }_0^{\mathrm{\infty }}\frac{x^i}{{(x+1)}^2}dx$$$$=\mathcal{R}e{\int }_0^{\mathrm{\infty }}\frac{x^{(i+1)-1}}{{(x+1)}^2}dx=\mathcal{R}e\left(\beta (1+i,1-i)\right)$$$$=\mathcal{R}e\frac{\mathrm{\Gamma }(1+i)\mathrm{\Gamma }(1-i)}{⟨\mathrm{\Gamma }\left(2\right)=1⟩}=$$$$\mathcal{R}e\left[i\mathrm{\Gamma }\left(i\right)\mathrm{\Gamma }(1-i)\right]\stackrel{eulerreflectionformula}{=}\mathcal{R}e\left[i\frac{\pi }{sin(i\pi }\right]$$$$=\mathcal{R}e\left[\frac{2i\left(i\pi \right)}{e^{i\left(i\pi \right)}-e^{-i\left(i\pi \right)}}\right]=\frac{-2\pi }{e^{-\pi }-e^{\pi }}$$$$=\frac{\pi }{\frac{e^{\pi }-e^{-\pi }}{2}}=\frac{\pi }{sinh\left(\pi \right)}✓$$$$m.n.july.1970$$$$$$

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