calvulate ∫_0 ^∞ ((cos(πx))/(1+x^4 ))dx


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Question Number 93633 by abdomathmax last updated on 14/May/20

$c a l v u l a t e \int_{0}^{\infty} \frac{c o s (π x)}{1 + x^{4}} d x$
Commented by mathmax by abdo last updated on 15/May/20
$A =∫_0 ^∞ ((cos(πx))/(x^4 +1))dx ⇒2A =∫_(−∞) ^(+∞) ((cos(πx))/(x^4 +1))dx =Re(∫_(−∞) ^(+∞) (e^(iπx) /(x^4 +1))dx) let ϕ(z) =(e^(iπz) /(z^4 +1)) poles of ϕ? ϕ(z) =(e^(iπz) /((z^2 −i)(z^2 +i))) =(e^(iπz) /((z−(√i))(z+(√i))(z−(√(−i)))(z+(√(−i))))) =(e^(iπz) /((z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{Res(ϕ,e^((iπ)/4) )+Res(ϕ,−e^(−((iπ)/4)) )} Res(ϕ,e^((iπ)/4) ) =(e^(iπ(e^((iπ)/4) )) /(2e^((iπ)/4) (2i))) =(1/(4i))e^(−((iπ)/4)) ×e^(iπ((1/(√2))+(i/(√2)))) =(1/(4i))e^(−((iπ)/4)) ×e^(−(π/(√2))) ×e^((iπ)/(√2)) =(e^(−(π/(√2))) /(4i)) e^(i((π/(√2))−(π/4))) Res(ϕ,−e^(−i(π/4)) ) = (e^(iπ (−e^(−((iπ)/4)) )) /(−2e^(−((iπ)/4)) (−2i))) =(1/(4i))e^((iπ)/4) e^(−iπ((1/(√2))−(i/(√2)))) =(1/(4i)) e^((iπ)/4) ×e^(π/(√2)) e^(−((iπ)/(√2))) =(e^(π/(√2)) /(4i)) ×e^(i((π/4)−(π/(√2)))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =(π/2){ e^(−(π/2)) ×e^(i((π/(√2))−(π/4))) +e^(π/2) × e^(i((π/4)−(π/(√2)))) } =((πe^(−(π/2)) )/2)( cos((π/(√2))−(π/4))+isin((π/(√2))−(π/4)))+((πe^(π/2) )/2)(cos((π/4)−(π/(√2)))+isin((π/4)−(π/(√2)))) =((π/2)e^(−(π/2)) +(π/2)e^(π/2) )cos((π/(√2))−(π/4)) +i(.....) ⇒ 2A =π ch((π/2))cos((π/(√2))−(π/4)) ⇒ A =(π/2)ch((π/2))cos((π/(√2))−(π/4))$
$A = \int_{0}^{\infty} \frac{c o s (π x)}{x^{4} + 1} d x \Rightarrow 2 A = \int_{- \infty}^{+ \infty} \frac{c o s (π x)}{x^{4} + 1} d x = R e (\int_{- \infty}^{+ \infty} \frac{e^{i π x}}{x^{4} + 1} d x)$ $l e t φ (z) = \frac{e^{i π z}}{z^{4} + 1} p o l e s o f φ ?$ $φ (z) = \frac{e^{i π z}}{(z^{2} - i) (z^{2} + i)} = \frac{e^{i π z}}{(z - \sqrt{i}) (z + \sqrt{i}) (z - \sqrt{- i}) (z + \sqrt{- i})}$ $= \frac{e^{i π z}}{(z - e^{\frac{i π}{4}}) (z + e^{\frac{i π}{4}}) (z - e^{- \frac{i π}{4}}) (z + e^{- \frac{i π}{4}})}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, e^{\frac{i π}{4}}) + R e s (φ, - e^{- \frac{i π}{4}})}$ $R e s (φ, e^{\frac{i π}{4}}) = \frac{e^{i π (e^{\frac{i π}{4}})}}{2 e^{\frac{i π}{4}} (2 i)} = \frac{1}{4 i} e^{- \frac{i π}{4}} \times e^{i π (\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}})}$ $= \frac{1}{4 i} e^{- \frac{i π}{4}} \times e^{- \frac{π}{\sqrt{2}}} \times e^{\frac{i π}{\sqrt{2}}} = \frac{e^{- \frac{π}{\sqrt{2}}}}{4 i} e^{i (\frac{π}{\sqrt{2}} - \frac{π}{4})}$ $R e s (φ, - e^{- i \frac{π}{4}}) = \frac{e^{i π (- e^{- \frac{i π}{4}})}}{- 2 e^{- \frac{i π}{4}} (- 2 i)} = \frac{1}{4 i} e^{\frac{i π}{4}} e^{- i π (\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}})}$ $= \frac{1}{4 i} e^{\frac{i π}{4}} \times e^{\frac{π}{\sqrt{2}}} e^{- \frac{i π}{\sqrt{2}}} = \frac{e^{\frac{π}{\sqrt{2}}}}{4 i} \times e^{i (\frac{π}{4} - \frac{π}{\sqrt{2}})} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = \frac{π}{2} {e^{- \frac{π}{2}} \times e^{i (\frac{π}{\sqrt{2}} - \frac{π}{4})} + e^{\frac{π}{2}} \times e^{i (\frac{π}{4} - \frac{π}{\sqrt{2}})}}$ $= \frac{π e^{- \frac{π}{2}}}{2} (c o s (\frac{π}{\sqrt{2}} - \frac{π}{4}) + i s i n (\frac{π}{\sqrt{2}} - \frac{π}{4})) + \frac{π e^{\frac{π}{2}}}{2} (c o s (\frac{π}{4} - \frac{π}{\sqrt{2}}) + i s i n (\frac{π}{4} - \frac{π}{\sqrt{2}}))$ $= (\frac{π}{2} e^{- \frac{π}{2}} + \frac{π}{2} e^{\frac{π}{2}}) c o s (\frac{π}{\sqrt{2}} - \frac{π}{4}) + i (. . . . .) \Rightarrow$ $2 A = π c h (\frac{π}{2}) c o s (\frac{π}{\sqrt{2}} - \frac{π}{4}) \Rightarrow A = \frac{π}{2} c h (\frac{π}{2}) c o s (\frac{π}{\sqrt{2}} - \frac{π}{4})$
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