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1-calculate-A-t-0-e-xt-sinxdx-with-x-gt-0-2-by-using-Fubuni-theorem-find-the-value-of-0-sinx-x-dx-

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Question Number 53967 by maxmathsup by imad last updated on 27/Jan/19
1)calculate A_t =∫_0 ^∞ e^(−xt) sinxdx with x>0 2) by using Fubuni theorem find the value of ∫_0 ^∞ ((sinx)/x)dx .
1)calculateAt=∫0∞e−xtsinxdxwithx>02)byusingFubunitheoremfindthevalueof∫0∞sinxxdx.
Commented by maxmathsup by imad last updated on 28/Jan/19
1) we have A_t =Im(∫_0 ^∞ e^(−xt) e^(ix) dx) =Im(∫_0 ^∞ e^((i−t)x) dx) ∫_0 ^∞ e^((i−t)x) dx =[(1/(i−x)) e^((i−t)x) ]_(x=0) ^(x=+∞) = −(1/(i−t)) =(1/(t−i)) =((t+i)/(t^2 +1)) ⇒ A_t = (1/(t^2 +1)) 2) we have ∫_0 ^∞ A_t dt =∫_0 ^∞ (dt/(t^2 +1)) =[arctant]_0 ^(+∞) =(π/2) and by fubini ∫_0 ^∞ A_t dt =∫_0 ^∞ (∫_0 ^∞ e^(−xt) sinxdx)dt =∫_0 ^∞ (∫_0 ^∞ e^(−xt) dt)sinxdx =∫_0 ^∞ ([−(1/x) e^(−xt) ]_(t=0) ^(t=+∞) )sinx dx =∫_0 ^∞ ((sinx)/x) dx ⇒ ∫_0 ^∞ ((sinx)/x) dx =(π/2) .
1)wehaveAt=Im(∫0∞e−xteixdx)=Im(∫0∞e(i−t)xdx)∫0∞e(i−t)xdx=[1i−xe(i−t)x]x=0x=+∞=−1i−t=1t−i=t+it2+1⇒At=1t2+12)wehave∫0∞Atdt=∫0∞dtt2+1=[arctant]0+∞=π2andbyfubini∫0∞Atdt=∫0∞(∫0∞e−xtsinxdx)dt=∫0∞(∫0∞e−xtdt)sinxdx=∫0∞([−1xe−xt]t=0t=+∞)sinxdx=∫0∞sinxxdx⇒∫0∞sinxxdx=π2.
Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jan/19
B_t =∫_0 ^∞ e^(−xt) cosxdx A_t =∫_0 ^∞ e^(−xt) sinxdx B_t +iA_t =∫_0 ^∞ e^(−xt) (cosx+isinx)dx B_t +iA_t =∫_0 ^∞ e^(−xt) .e^(ix) dx=∫_0 ^∞ e^(−xt+ix) dx =∫_0 ^∞ e^(x(−t+i)) dx=∣(e^(x(−t+i)) /(−t+i))∣_0 ^∞ =∣(e^(−x(t−i)) /(−t+i))∣_0 ^∞ =((e^(−∞(t−i)) −e^0 )/(−t+i))=((−1)/(−t+i))=(1/(t−i)) =((t+i)/(t^2 +1))=(t/(t^2 +1))+i×(1/(t^2 +1)) so B_t =(t/(t^2 +1)) A_t =(1/(t^2 +1))
Bt=∫0∞e−xtcosxdxAt=∫0∞e−xtsinxdxBt+iAt=∫0∞e−xt(cosx+isinx)dxBt+iAt=∫0∞e−xt.eixdx=∫0∞e−xt+ixdx=∫0∞ex(−t+i)dx=∣ex(−t+i)−t+i∣0∞=∣e−x(t−i)−t+i∣0∞=e−∞(t−i)−e0−t+i=−1−t+i=1t−i=t+it2+1=tt2+1+i×1t2+1soBt=tt2+1At=1t2+1
Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jan/19
C_t =∫_0 ^∞ e^(−xt) ((sinx)/x)dx (dC_t /dt)=∫_0 ^∞ (∂/∂t)(((e^(−xt) sinx)/x))dx =∫_0 ^∞ ((e^(−xt) ×−x×sinx)/x)dx =−∫_0 ^∞ e^(−xt) sinxdx=−A_t (dC_t /dt)=−(1/(t^2 +1)) −dC_t =(dt/(t^2 +1)) −C_t =tan^(−1) (t)+k k=−C_t −tan^(−1) (t) when t→∞ C_t →0 and tan^(−1) (t)→(π/2) so k=−(π/2) −C_t =tan^(−1) (t)−(π/2) we have to find ∫_0 ^∞ ((sinx)/x)dx we know that −∫_0 ^∞ e^(−xt) ((sinx)/x)dx=tan^(−1) (t)−(π/2) now put t=0 botb side −∫_0 ^∞ ((sinx)/x)=tan^(−1) (0)−(π/2) so ∫_0 ^∞ ((sinx)/x)=(π/2) proved
Ct=∫0∞e−xtsinxxdxdCtdt=∫0∞∂∂t(e−xtsinxx)dx=∫0∞e−xt×−x×sinxxdx=−∫0∞e−xtsinxdx=−AtdCtdt=−1t2+1−dCt=dtt2+1−Ct=tan−1(t)+kk=−Ct−tan−1(t)whent→∞Ct→0andtan−1(t)→π2sok=−π2−Ct=tan−1(t)−π2wehavetofind∫0∞sinxxdxweknowthat−∫0∞e−xtsinxxdx=tan−1(t)−π2nowputt=0botbside−∫0∞sinxx=tan−1(0)−π2so∫0∞sinxx=π2proved

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