Menu Close

prove-that-0-cos-x-chx-dx-2-n-0-1-n-2n-1-2n-1-2-2-

Tinku Tara 0 Comments
Pin




Question Number 33350 by caravan msup abdo. last updated on 14/Apr/18
prove that ∫_0 ^∞ ((cos(αx))/(chx))dx= 2 Σ_(n=0) ^∞ (−1)^n ((2n+1)/((2n+1)^2 +α^2 )) .
provethat∫0∞cos(αx)chxdx=2∑n=0∞(−1)n2n+1(2n+1)2+α2.
Commented by abdo imad last updated on 17/Apr/18
let put I =∫_0 ^∞ ((cos(αx))/(chx))dx I = 2 ∫_0 ^∞ ((cos(αx))/(e^x +e^(−x) ))dx =2 ∫_0 ^∞ ((e^(−x) cos(αx))/(1+e^(−2x) ))dx =2 ∫_0 ^∞ (Σ_(n=0) ^∞ (−1)^n e^(−2nx) ) .e^(−x) cos(αx)dx =2 Σ_(n=0) ^∞ (−1)^n ∫_0 ^∞ e^(−(2n+1)x) cos(αx)dx but A_n =∫_0 ^∞ e^(−(2n+1)x) cos(αx)dx = Re( ∫_0 ^∞ e^(−(2n+1)x) e^(−iαx) dx) =Re( ∫_0 ^∞ e^(−(2n+1+iα)x) dx) but we have ∫_0 ^∞ e^(−(2n+1 +iα)x) dx =[−(1/(2n+1 +iα)) e^(−(2n+1 +iα)) ]_0 ^(+∞) =(1/(2n+1+iα)) = ((2n+1−iα)/((2n+1)^2 +α^2 )) ⇒ A_n =((2n+1)/((2n+1)^2 +α^2 )) ⇒ I =2 Σ_(n=0) ^∞ (((−1)^n (2n+1))/((2n+1)^2 +α^2 )) .
letputI=∫0∞cos(αx)chxdxI=2∫0∞cos(αx)ex+e−xdx=2∫0∞e−xcos(αx)1+e−2xdx=2∫0∞(∑n=0∞(−1)ne−2nx).e−xcos(αx)dx=2∑n=0∞(−1)n∫0∞e−(2n+1)xcos(αx)dxbutAn=∫0∞e−(2n+1)xcos(αx)dx=Re(∫0∞e−(2n+1)xe−iαxdx)=Re(∫0∞e−(2n+1+iα)xdx)butwehave∫0∞e−(2n+1+iα)xdx=[−12n+1+iαe−(2n+1+iα)]0+∞=12n+1+iα=2n+1−iα(2n+1)2+α2⇒An=2n+1(2n+1)2+α2⇒I=2∑n=0∞(−1)n(2n+1)(2n+1)2+α2.

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *