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Question Number 75986 by Ajao yinka last updated on 21/Dec/19

Commented by abdomathmax last updated on 24/Dec/19

let f(t) =∫_0 ^∞ ((arctan(xt)cos(nx))/x)dx with t≥0 f^′ (t)=∫_0 ^∞ (x/(1+x^2 t^2 ))((cos(nx))/x)dx =∫_0 ^∞ ((cos(nx))/(x^2 t^2 +1))dx =_(xt=u) ∫_0 ^∞ ((cos(n(u/t)))/(u^2 +1))(du/t) =(1/(2t)) ∫_(−∞) ^(+∞) ((cos(((nu)/t)))/(u^2 +1)) du =(1/(2t)) Re( ∫_(−∞) ^(+∞) (e^(i(((nu)/t))) /(u^2 +1))du) let W(z) =(e^(i(((nz)/t))) /(z^2 +1)) we hsve? ∫_(−∞) ^(+∞) W(z)dz =2iπ Res(W,i) =2iπ ×(e^(−(n/t)) /(2i)) =π e^(−(n/t)) ⇒f^′ (t) =(π/(2t)) e^(−(n/t)) ⇒ f(t) =(π/2)∫_1 ^t (e^(−(n/z)) /z)dz +c ....be continued...

$${let}\:{f}\left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({xt}\right){cos}\left({nx}\right)}{{x}}{dx}\:\:{with}\:{t}\geqslant\mathrm{0} \\ $$$${f}^{'} \left({t}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{2}} }\frac{{cos}\left({nx}\right)}{{x}}{dx}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({nx}\right)}{{x}^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$$$=_{{xt}={u}} \:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({n}\frac{{u}}{{t}}\right)}{{u}^{\mathrm{2}} \:+\mathrm{1}}\frac{{du}}{{t}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{t}}\:\int_{−\infty} ^{+\infty} \:\frac{{cos}\left(\frac{{nu}}{{t}}\right)}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:{du}\:=\frac{\mathrm{1}}{\mathrm{2}{t}}\:{Re}\left(\:\int_{−\infty} ^{+\infty} \:\frac{{e}^{{i}\left(\frac{{nu}}{{t}}\right)} }{{u}^{\mathrm{2}} \:+\mathrm{1}}{du}\right) \\ $$$${let}\:{W}\left({z}\right)\:=\frac{{e}^{{i}\left(\frac{{nz}}{{t}}\right)} }{{z}^{\mathrm{2}} \:+\mathrm{1}}\:\:{we}\:{hsve}? \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({W},{i}\right)\:=\mathrm{2}{i}\pi\:×\frac{{e}^{−\frac{{n}}{{t}}} }{\mathrm{2}{i}} \\ $$$$=\pi\:{e}^{−\frac{{n}}{{t}}} \:\Rightarrow{f}^{'} \left({t}\right)\:=\frac{\pi}{\mathrm{2}{t}}\:{e}^{−\frac{{n}}{{t}}} \:\Rightarrow \\ $$$${f}\left({t}\right)\:=\frac{\pi}{\mathrm{2}}\int_{\mathrm{1}} ^{{t}} \:\:\:\frac{{e}^{−\frac{{n}}{{z}}} }{{z}}{dz}\:+{c}\:....{be}\:{continued}... \\ $$

Answered by mind is power last updated on 23/Dec/19

f(t)=∫_0 ^(+∞) ((tan^− (xt)cos(nx))/x)dx f′(t)=∫_0 ^(+∞) ((cos(nx))/(1+x^2 t^2 ))dx =n∫_0 ^(+∞) ((cos(z))/(n^2 +z^2 t^2 )).dz =(n/2)∫_(−∞) ^(+∞) ((cos(z))/(n^2 +z^2 t^2 ))dz =Re((n/2)∫_(−∞) ^(+∞) ((e^(iz) )/(n^2 +z^2 t^2 ))dz) by residu th f(z)=(e^(iz) /(n^2 +z^2 t^2 )) =(n/2).2iπ{Res(f,((ni)/t)))} inπ(e^(−(n/t)) /(2int))}=(π/(2t))e^(−(n/t)) f(0)=∫_0 ^(+∞) ((arvtan(0)cos(nx)dx)/x)=0 f(1)=Δ=∫_0 ^1 f′(t)dt=∫_0 ^1 (π/(2t))(e^((−n)/t) )dt=(π/2)∫_0 ^1 (e^(−(n/t)) /t)dt w=(n/t)⇒dt=−n(dw/w^2 ) Δ=−((nπ)/2)∫_1 ^(+∞) (e^(−w) /w)dw Ei(x)=−∫_(−x) ^(+∞) (e^(−t) /t)dt ,x<0 Δ=−((nπ)/2)E_i (−1)

$$\mathrm{f}\left(\mathrm{t}\right)=\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{tan}^{−} \left(\mathrm{xt}\right)\mathrm{cos}\left(\mathrm{nx}\right)}{\mathrm{x}}\mathrm{dx} \\ $$$$\mathrm{f}'\left(\mathrm{t}\right)=\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{cos}\left(\mathrm{nx}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} \mathrm{t}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\mathrm{n}\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{cos}\left(\mathrm{z}\right)}{\mathrm{n}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} \mathrm{t}^{\mathrm{2}} }.\mathrm{dz} \\ $$$$=\frac{\mathrm{n}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \frac{\mathrm{cos}\left(\mathrm{z}\right)}{\mathrm{n}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} \mathrm{t}^{\mathrm{2}} }\mathrm{dz}\:=\mathrm{Re}\left(\frac{\mathrm{n}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \frac{\mathrm{e}^{\mathrm{iz}} \:}{\mathrm{n}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} \mathrm{t}^{\mathrm{2}} }\mathrm{dz}\right)\:\mathrm{by}\:\mathrm{residu}\:\mathrm{th} \\ $$$$\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{iz}} }{\mathrm{n}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} \mathrm{t}^{\mathrm{2}} } \\ $$$$\left.=\frac{\mathrm{n}}{\mathrm{2}}.\mathrm{2i}\pi\left\{\mathrm{Res}\left(\mathrm{f},\frac{\mathrm{ni}}{\mathrm{t}}\right)\right)\right\} \\ $$$$\left.\mathrm{in}\pi\frac{\mathrm{e}^{−\frac{\mathrm{n}}{\mathrm{t}}} }{\mathrm{2int}}\right\}=\frac{\pi}{\mathrm{2t}}\mathrm{e}^{−\frac{\mathrm{n}}{\mathrm{t}}} \\ $$$$\mathrm{f}\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{arvtan}\left(\mathrm{0}\right)\mathrm{cos}\left(\mathrm{nx}\right)\mathrm{dx}}{\mathrm{x}}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{f}\left(\mathrm{1}\right)=\Delta=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{f}'\left(\mathrm{t}\right)\mathrm{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\pi}{\mathrm{2t}}\left(\mathrm{e}^{\frac{−\mathrm{n}}{\mathrm{t}}} \right)\mathrm{dt}=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{e}^{−\frac{\mathrm{n}}{\mathrm{t}}} }{\mathrm{t}}\mathrm{dt} \\ $$$$\mathrm{w}=\frac{\mathrm{n}}{\mathrm{t}}\Rightarrow\mathrm{dt}=−\mathrm{n}\frac{\mathrm{dw}}{\mathrm{w}^{\mathrm{2}} } \\ $$$$\Delta=−\frac{\mathrm{n}\pi}{\mathrm{2}}\int_{\mathrm{1}} ^{+\infty} \frac{\mathrm{e}^{−\mathrm{w}} }{\mathrm{w}}\mathrm{dw} \\ $$$$\mathrm{Ei}\left(\mathrm{x}\right)=−\int_{−\mathrm{x}} ^{+\infty} \frac{\mathrm{e}^{−\mathrm{t}} }{\mathrm{t}}\mathrm{dt}\:\:\:\:\:,\mathrm{x}<\mathrm{0} \\ $$$$\Delta=−\frac{\mathrm{n}\pi}{\mathrm{2}}\mathrm{E}_{\mathrm{i}} \left(−\mathrm{1}\right) \\ $$$$ \\ $$$$ \\ $$

Commented by Ajao yinka last updated on 31/Dec/19

Perfect

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