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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...

letf(t)=0arctan(xt)cos(nx)xdxwitht0f(t)=0x1+x2t2cos(nx)xdx=0cos(nx)x2t2+1dx=xt=u0cos(nut)u2+1dut=12t+cos(nut)u2+1du=12tRe(+ei(nut)u2+1du)letW(z)=ei(nzt)z2+1wehsve?+W(z)dz=2iπRes(W,i)=2iπ×ent2i=πentf(t)=π2tentf(t)=π21tenzzdz+c....becontinued...

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)

f(t)=0+tan(xt)cos(nx)xdxf(t)=0+cos(nx)1+x2t2dx=n0+cos(z)n2+z2t2.dz=n2+cos(z)n2+z2t2dz=Re(n2+eizn2+z2t2dz)byresiduthf(z)=eizn2+z2t2=n2.2iπ{Res(f,nit))}inπent2int}=π2tentf(0)=0+arvtan(0)cos(nx)dxx=0f(1)=Δ=01f(t)dt=01π2t(ent)dt=π201enttdtw=ntdt=ndww2Δ=nπ21+ewwdwEi(x)=x+ettdt,x<0Δ=nπ2Ei(1)

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

Perfect

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