Question-12422

Question Number 12422 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 21/Apr/17

Commented by mrW1 last updated on 22/Apr/17

f(x)=∫(dx/(x^2 −2xcos ϕ+1))=F(x,ϕ)+C f(0)=F(0,ϕ)+C lim_(ϕ→0) f(0)=lim_(ϕ→0) F(0,ϕ)+C due to the constant C I think none of the given answers is true.

$${f}\left({x}\right)=\int\frac{{dx}}{{x}^{\mathrm{2}} −\mathrm{2}{x}\mathrm{cos}\:\varphi+\mathrm{1}}={F}\left({x},\varphi\right)+{C} \\ $$$${f}\left(\mathrm{0}\right)={F}\left(\mathrm{0},\varphi\right)+{C} \\ $$$$\underset{\varphi\rightarrow\mathrm{0}} {\mathrm{lim}}\:{f}\left(\mathrm{0}\right)=\underset{\varphi\rightarrow\mathrm{0}} {\mathrm{lim}}\:{F}\left(\mathrm{0},\varphi\right)+{C} \\ $$$$ \\ $$$${due}\:{to}\:{the}\:{constant}\:{C}\:{I}\:{think}\:{none} \\ $$$${of}\:{the}\:{given}\:{answers}\:{is}\:{true}. \\ $$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/Apr/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/Apr/17

f(x)=(1/(sinϕ))arctg(((x−cosϕ)/(sinϕ)))+C lim_(ϕ→0) f(0)=lim_(ϕ→0) (1/(sinϕ))arctg(((−cosϕ)/(sinϕ)))+C= =lim_(ϕ→0) (1/(sinϕ))arctg(tg(90+ϕ))+C=lim_(ϕ→0) ((90+ϕ)/(sinϕ))+C= =lim_(ϕ→0) ((90+ϕ+Csinϕ)/(sinϕ))=1+C. mrW1!you are right.we can not determine the lim_(ϕ→0) f(0). but this integral is so beautiful.

$${f}\left({x}\right)=\frac{\mathrm{1}}{{sin}\varphi}{arctg}\left(\frac{{x}−{cos}\varphi}{{sin}\varphi}\right)+{C} \\ $$$$\underset{\varphi\rightarrow\mathrm{0}} {\mathrm{lim}}{f}\left(\mathrm{0}\right)=\underset{\varphi\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{sin}\varphi}{arctg}\left(\frac{−{cos}\varphi}{{sin}\varphi}\right)+{C}= \\ $$$$=\underset{\varphi\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{sin}\varphi}{arctg}\left({tg}\left(\mathrm{90}+\varphi\right)\right)+{C}=\underset{\varphi\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{90}+\varphi}{{sin}\varphi}+{C}= \\ $$$$=\underset{\varphi\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{90}+\varphi+{Csin}\varphi}{{sin}\varphi}=\mathrm{1}+{C}. \\ $$$${mrW}\mathrm{1}!{you}\:{are}\:{right}.{we}\:{can}\:{not}\: \\ $$$${determine}\:{the}\:\underset{\varphi\rightarrow\mathrm{0}} {\mathrm{lim}}{f}\left(\mathrm{0}\right). \\ $$$${but}\:{this}\:{integral}\:{is}\:{so}\:{beautiful}. \\ $$

Commented by mrW1 last updated on 22/Apr/17

I=F(x,ϕ)+C F(x,ϕ)=(1/(sin ϕ))×tan^(−1) (((x−cos ϕ)/(sin ϕ))) F(0,ϕ)=(1/(sin ϕ))×tan^(−1) (((−cos ϕ)/(sin ϕ))) =−(1/(sin ϕ))×tan^(−1) [tan ((π/2)−ϕ)] =((ϕ−(π/2))/(sin ϕ)) f(0)=F(0,ϕ)=((ϕ−(π/2))/(sin ϕ))+C lim_(ϕ→0) f(0)=1+∞+C=∞

$${I}={F}\left({x},\varphi\right)+{C} \\ $$$${F}\left({x},\varphi\right)=\frac{\mathrm{1}}{\mathrm{sin}\:\varphi}×\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}−\mathrm{cos}\:\varphi}{\mathrm{sin}\:\varphi}\right) \\ $$$${F}\left(\mathrm{0},\varphi\right)=\frac{\mathrm{1}}{\mathrm{sin}\:\varphi}×\mathrm{tan}^{−\mathrm{1}} \left(\frac{−\mathrm{cos}\:\varphi}{\mathrm{sin}\:\varphi}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{sin}\:\varphi}×\mathrm{tan}^{−\mathrm{1}} \left[\mathrm{tan}\:\left(\frac{\pi}{\mathrm{2}}−\varphi\right)\right] \\ $$$$=\frac{\varphi−\frac{\pi}{\mathrm{2}}}{\mathrm{sin}\:\varphi} \\ $$$${f}\left(\mathrm{0}\right)={F}\left(\mathrm{0},\varphi\right)=\frac{\varphi−\frac{\pi}{\mathrm{2}}}{\mathrm{sin}\:\varphi}+{C} \\ $$$$\underset{\varphi\rightarrow\mathrm{0}} {\mathrm{lim}}\:{f}\left(\mathrm{0}\right)=\mathrm{1}+\infty+{C}=\infty \\ $$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/Apr/17