Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 98678 by PRITHWISH SEN 2 last updated on 15/Jun/20

$${\int }_0^{\mathrm{\infty }}{\mathbf{e}}^{-\mathbf{ax}}\frac{\sin \mathbf{mx}}{\mathbf{x}}\mathbf{dx}={\tan }^{-1}\left(\frac{\mathbf{m}}{\mathbf{a}}\right),\mathbf{a}>0$$

Commented byTinku Tara last updated on 15/Jun/20

$$\mathrm{Are}\mathrm{you}\mathrm{facing}\mathrm{problem}\mathrm{with}$$ $$2.083?\mathrm{please}\mathrm{let}\mathrm{us}\mathrm{know}.\mathrm{if}$$ $$\mathrm{there}\mathrm{is}\mathrm{crash}\mathrm{please}\mathrm{report}.$$

Commented byPRITHWISH SEN 2 last updated on 15/Jun/20

$$\mathrm{actually}\mathrm{it}\mathrm{is}\mathrm{not}\mathrm{opening}\mathrm{for}\mathrm{most}\mathrm{of}\mathrm{the}\mathrm{time}.$$

Commented byTinku Tara last updated on 15/Jun/20

$$\mathrm{Do}\mathrm{u}\mathrm{get}\mathrm{a}\mathrm{crash}\mathrm{or}\mathrm{something}?$$

Commented byTinku Tara last updated on 15/Jun/20

$$\mathrm{Like}\mathrm{do}\mathrm{u}\mathrm{get}{}^{\prime }\mathrm{App}{\mathrm{stopped}}^{\prime }$$

Commented byRasheed.Sindhi last updated on 15/Jun/20

$$Whereisthisversion?$$ $$www.tinkutara.comoffers$$ $$2.081aslattestversion!$$

Commented byTinku Tara last updated on 15/Jun/20

$$\mathrm{Playstore}\mathrm{and}\mathrm{tinkutara}.\mathrm{com}$$ $$\mathrm{both}\mathrm{have}2.083.2.083\mathrm{introduced}$$ $$\mathrm{background}\mathrm{color}\mathrm{and}\mathrm{more}\mathrm{font}$$ $$\mathrm{color}.$$

Commented byRasheed.Sindhi last updated on 15/Jun/20

$$\mathcal{H}owtochangebackgroundsir?$$

Commented byTinku Tara last updated on 15/Jun/20

$$\mathrm{only}\mathrm{available}\mathrm{in}\mathrm{v}2.083\mathrm{in}\mathrm{offline}$$ $$\mathrm{editor}.$$

Commented byPRITHWISH SEN 2 last updated on 16/Jun/20

$$\mathrm{I}\mathrm{have}\mathrm{updated}2.084\mathrm{version}.\mathrm{It}\mathrm{is}\mathrm{opening}\mathrm{but}\mathrm{it}$$ $$\mathrm{is}\mathrm{taking}\mathrm{some}\mathrm{time}.$$

Answered by mathmax by abdo last updated on 15/Jun/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{ax}}\frac{\sin \left(\mathrm{mx}\right)}{\mathrm{x}}\mathrm{dx}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{ax}}\sin \left(\mathrm{mx}\right)\mathrm{dx}$$ $$=-\mathrm{Im}\left({\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{ax}+\mathrm{imx}}\mathrm{dx}\right)\mathrm{we}\mathrm{have}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{(-\mathrm{a}+\mathrm{im})\mathrm{x}}\mathrm{dx}={\left[\frac{1}{-\mathrm{a}+\mathrm{im}}{\mathrm{e}}^{(-\mathrm{a}+\mathrm{im})\mathrm{x}}\right]}_0^{\mathrm{\infty }}$$ $$=-\frac{1}{\mathrm{a}-\mathrm{im}}(-1)=\frac{1}{\mathrm{a}-\mathrm{im}}=\frac{\mathrm{a}+\mathrm{im}}{{\mathrm{a}}^2+{\mathrm{m}}^2}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-\frac{\mathrm{m}}{{\mathrm{a}}^2+{\mathrm{m}}^2}\Rightarrow$$ $$\mathrm{f}\left(\mathrm{a}\right)=-\mathrm{m}\int \frac{\mathrm{da}}{{\mathrm{a}}^2+{\mathrm{m}}^2}+\mathrm{c}(\mathrm{a}=\mathrm{mu})$$ $$=-\mathrm{m}\int \frac{\mathrm{mdu}}{{\mathrm{m}}^2(1+{\mathrm{u}}^2)}=-\int \frac{\mathrm{du}}{1+{\mathrm{u}}^2}=-\arctan \left(\frac{\mathrm{a}}{\mathrm{m}}\right)+\mathrm{c}(\mathrm{we}\mathrm{suppoze}\mathrm{m}\ne 0)$$ $$\mathrm{if}\mathrm{m}>0\mathrm{we}\mathrm{get}{\lim }_{\mathrm{a}\to +\mathrm{\infty }}\mathrm{f}\left(\mathrm{a}\right)=0=-\frac{\pi }{2}+\mathrm{c}\Rightarrow \mathrm{c}=\frac{\pi }{2}\Rightarrow$$ $$\mathrm{f}\left(\mathrm{a}\right)=\frac{\pi }{2}-\arctan \left(\frac{\mathrm{a}}{\mathrm{m}}\right)=\arctan \left(\frac{\mathrm{m}}{\mathrm{a}}\right)$$ $$\mathrm{if}\mathrm{m}<0\mathrm{we}\mathrm{get}{\lim }_{\mathrm{a}\to +\mathrm{\infty }}\mathrm{f}\left(\mathrm{a}\right)=0=\frac{\pi }{2}+\mathrm{c}\Rightarrow \mathrm{c}=-\frac{\pi }{2}\Rightarrow$$ $$\mathrm{f}\left(\mathrm{a}\right)=-\arctan \left(\frac{\mathrm{a}}{\mathrm{m}}\right)-\frac{\pi }{2}=-(\frac{\pi }{2}+\mathrm{arctam}\left(\frac{\mathrm{a}}{\mathrm{m}}\right))$$ $$=-(\frac{\pi }{2}-\arctan (-\frac{\mathrm{a}}{\mathrm{m}}))=-\arctan (-\frac{\mathrm{m}}{\mathrm{a}})=\arctan \left(\frac{\mathrm{m}}{\mathrm{a}}\right)\mathrm{so}\mathrm{in}\mathrm{all}\mathrm{case}$$ $$\mathrm{we}\mathrm{have}\mathrm{f}\left(\mathrm{a}\right)=\arctan \left(\frac{\mathrm{m}}{\mathrm{a}}\right)$$ $$\mathrm{if}\mathrm{m}=0\Rightarrow \mathrm{f}\left(\mathrm{a}\right)=0$$ $$$$

Commented byPRITHWISH SEN 2 last updated on 15/Jun/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented bymathmax by abdo last updated on 15/Jun/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

Answered by smridha last updated on 15/Jun/20

$${\mathit{l}\mathit{e}\mathit{t}}^{\prime }s\mathit{d}\mathit{o}\mathit{i}\mathit{t}\mathit{b}\mathit{y}\mathit{L}aplace\mathit{t}\mathit{r}\mathit{a}\mathit{n}\mathit{s}\mathit{f}\mathit{o}\mathit{r}\mathit{m}$$ $$\mathit{f}\left(\mathit{s}\right)=\mathit{L}\left[\mathit{s}\mathit{i}\mathit{n}\left(\mathit{m}\mathit{x}\right)\right]={\int }_0^{\mathrm{\infty }}{\mathit{e}}^{-\mathit{s}\mathit{x}}\mathit{s}\mathit{i}\mathit{n}\left(\mathit{m}\mathit{x}\right)\mathit{d}\mathit{x}$$ $$=\frac{\mathit{m}}{{\mathit{s}}^2+{\mathit{m}}^2}$$ $$\mathit{n}\mathit{o}\mathit{w}\mathit{L}\left[\frac{\mathit{s}\mathit{i}\mathit{n}\left(\mathit{m}\mathit{x}\right)}{\mathit{x}}\right]={\int }_{\mathit{a}}^{\mathrm{\infty }}\frac{\mathit{m}}{{\mathit{s}}^2+{\mathit{m}}^2}\mathit{d}\mathit{s}$$ $$=\mathit{m}.\frac{1}{\mathit{m}}{\left[{\tan }^{-1}\left(\frac{\mathit{s}}{\mathit{m}}\right)\right]}_{\mathit{a}}^{\mathrm{\infty }}$$ $$=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{\mathit{a}}{\mathit{m}}\right)={\cot }^{-1}\left(\frac{\mathit{a}}{\mathit{m}}\right)={\tan }^{-1}\left(\frac{\mathit{m}}{\mathit{a}}\right)\mathit{w}\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathit{a}>0$$

Commented byPRITHWISH SEN 2 last updated on 16/Jun/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented bysmridha last updated on 16/Jun/20

$$\mathit{w}\mathit{e}\mathit{l}\mathit{c}\mathit{o}\mathit{m}\mathit{e}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com