calculate ∫_0 ^∞ (dx/((x^2 −e^(ia) )(x^2 −e^(ib) ))) with a>0 andb>0


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Question Number 67799 by mathmax by abdo last updated on 31/Aug/19

$c a l c u l a t e \int_{0}^{\infty} \frac{d x}{(x^{2} - e^{i a}) (x^{2} - e^{i b})} w i t h a > 0 a n d b > 0$
Commented by MJS last updated on 31/Aug/19

$can we calculate \underset{0}{\int^{\infty}} \frac{d x}{(x^{2} - α) (x^{2} - β)} and then$ $insert α = e^{i a}; β = e^{i b} ?$
Commented by mathmax by abdo last updated on 31/Aug/19

$y e s s i r b u t y o u m u s t d e t e r m i n e \int \frac{d x}{x^{2} - z} w i t h z f r o m C .$
Commented by mathmax by abdo last updated on 01/Sep/19
$let I =∫_0 ^∞ (dx/((x^2 −e^(ia) )(x^2 −e^(ib) ))) ⇒2I =∫_(−∞) ^(+∞) (dx/((x^2 −e^(ia) )(x^2 −e^(ib) ))) let ϕ(z) =(1/((z^2 −e^(ia) )(z^2 −e^(ib) ))) ⇒ϕ(z) =(1/((z−e^((ia)/2) )(z+e^((ia)/2) )(z−e^((ib)/2) )(z+e^((ib)/2) ))) the poles of ϕ are +^− e^((ia)/2) and +^− e^((ib)/(2 )) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,e^((ia)/2) )+Res(ϕ,e^((ib)/2) )} Res(ϕ,e^((ia)/2) )=lim_(z→e^((ia)/2) ) (z−e^((ia)/2) )ϕ(z) =(1/(2e^((ia)/2) (e^(ia) −e^(ib) ))) =(e^(−i(a/2)) /(2(e^(ia) −e^(ib) ))) Res(ϕ,e^((ib)/2) ) =lim_(z→e^((ib)/2) ) (z−e^((ib)/2) )ϕ(z) =(e^(−((ib)/2)) /(2(e^(ib) −e^(ia) ))) ⇒ ∫_(−∞) ^(+∞) ϕ(2)dz =2iπ{(e^(−((ia)/2)) /(2(e^(ia) −e^(ib) ))) +(e^(−((ib)/2)) /(2(e^(ib) −e^(ia) )))} =((iπ)/(e^(ia) −e^(ib) )){ e^(−((ia)/2)) −e^(−((ib)/2)) } =2I ⇒ I =((iπ(e^(−((ia)/2)) −e^(−((ib)/2)) ))/(2(e^(ia) −e^(ib) ))).$
$l e t I = \int_{0}^{\infty} \frac{d x}{(x^{2} - e^{i a}) (x^{2} - e^{i b})} \Rightarrow 2 I = \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} - e^{i a}) (x^{2} - e^{i b})}$ $l e t φ (z) = \frac{1}{(z^{2} - e^{i a}) (z^{2} - e^{i b})} \Rightarrow φ (z) = \frac{1}{(z - e^{\frac{i a}{2}}) (z + e^{\frac{i a}{2}}) (z - e^{\frac{i b}{2}}) (z + e^{\frac{i b}{2}})}$ $t h e p o l e s o f φ a r e \bar{+} e^{\frac{i a}{2}} a n d \bar{+} e^{\frac{i b}{2}} r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, e^{\frac{i a}{2}}) + R e s (φ, e^{\frac{i b}{2}})}$ $R e s (φ, e^{\frac{i a}{2}}) = {l i m}_{z \to e^{\frac{i a}{2}}} (z - e^{\frac{i a}{2}}) φ (z) = \frac{1}{2 e^{\frac{i a}{2}} (e^{i a} - e^{i b})} = \frac{e^{- i \frac{a}{2}}}{2 (e^{i a} - e^{i b})}$ $R e s (φ, e^{\frac{i b}{2}}) = {l i m}_{z \to e^{\frac{i b}{2}}} (z - e^{\frac{i b}{2}}) φ (z) = \frac{e^{- \frac{i b}{2}}}{2 (e^{i b} - e^{i a})} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (2) d z = 2 i π {\frac{e^{- \frac{i a}{2}}}{2 (e^{i a} - e^{i b})} + \frac{e^{- \frac{i b}{2}}}{2 (e^{i b} - e^{i a})}}$ $= \frac{i π}{e^{i a} - e^{i b}} {e^{- \frac{i a}{2}} - e^{- \frac{i b}{2}}} = 2 I \Rightarrow$ $I = \frac{i π (e^{- \frac{i a}{2}} - e^{- \frac{i b}{2}})}{2 (e^{i a} - e^{i b})} .$
	Answered by mind is power last updated on 01/Sep/19
	$let γ_r ={z∈C such that∣z∣≤r ∣im(z)>0} we have ∫_0 ^(+∞) (dx/((x^2 −e^(ia) )(x^2 −e^(ib) )))=(1/2)∫_(−∞) ^(+∞) (dx/((x^2 −e^(ia) )(x^2 −e^(ib) ))) let f(z)=(1/((z^2 −e^(ia) )(z^2 −e^(ib) ))) f(z)=(1/((r^2 e^(2i∅) −e^(ia) )(r^2 e^(2i∅) −e^(ib) ))) over γ_r ∣f(z)∣≤(1/(∣r^2 −1∣^2 )) we used ∣z+z′∣≥∣∣z∣−∣z′∣∣ twice ⇒∫_(γr) f(z)dz≤∫_γ_r (1/((r^2 −1)^2 ))dz=((πr)/((r^2 −1)^2 ))→0=as r→+∞ residu theorem poles of f inside γ_r ∪]−∞.+∞[ ar e^(i(a/2)) .e^((ib)/2) lim (z−e^(i(a/2)) ).f(z)=(1/(2e^(i(a/2)) .(e^(ia) −e^(ib) ))) lim(z−e^(i(b/2)) )f(z)=(1/(2e^(i(b/2)) (e^(ib) −e^(ia) ))) ∫_(−∞) ^(+∞) f(z)dz=2iπΣ_(residu) f(z)=2iπ((e^(−i(b/2)) /((e^(ib) −e^(ia) )2))−(e^(−((ia)/2)) /(2(e^(ib) −e^(ia) )))) =iπ(((e^(−((ib)/2)) −e^((−ia)/2) )/((e^(i((a+b)/2)) (e^(i((b−a)/2)) −e^(i−((b−a)/2)) ))))=iπ((e^(−ib−i(a/2)) −e^(−ia−i(b/2)) )/(2isin(((a−b)/2))))$
	$l e t γ_{r} = {z \in C s u c h t h a t ∣ z ∣⩽ r ∣ i m (z) > 0}$ $w e h a v e \int_{0}^{+ \infty} \frac{d x}{(x^{2} - e^{i a}) (x^{2} - e^{i b})} = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} - e^{i a}) (x^{2} - e^{i b})}$ $l e t f (z) = \frac{1}{(z^{2} - e^{i a}) (z^{2} - e^{i b})}$ $f (z) = \frac{1}{(r^{2} e^{2 i \emptyset} - e^{i a}) (r^{2} e^{2 i \emptyset} - e^{i b})} o v e r γ_{r}$ $∣ f (z) ∣⩽ \frac{1}{∣ r^{2} - 1 ∣^{2}}$ $w e u s e d ∣ z + z^{'} ∣⩾∣∣ z ∣ - ∣ z^{'} ∣∣ t w i c e$ $\Rightarrow \int_{γ r} f (z) d z ⩽ \int_{γ_{r}} \frac{1}{{(r^{2} - 1)}^{2}} d z = \frac{π r}{{(r^{2} - 1)}^{2}} \to 0 = a s r \to + \infty$ $r e s i d u t h e o r e m$ $p o l e s o f f i n s i d e γ_{r} \cup] - \infty . + \infty [a r e^{i \frac{a}{2}} . e^{\frac{i b}{2}}$ $l i m (z - e^{i \frac{a}{2}}) . f (z) = \frac{1}{2 e^{i \frac{a}{2}} . (e^{i a} - e^{i b})}$ $l i m (z - e^{i \frac{b}{2}}) f (z) = \frac{1}{2 e^{i \frac{b}{2}} (e^{i b} - e^{i a})}$ $\int_{- \infty}^{+ \infty} f (z) d z = 2 i π \sum_{r e s i d u} f (z) = 2 i π (\frac{e^{- i \frac{b}{2}}}{(e^{i b} - e^{i a}) 2} - \frac{e^{- \frac{i a}{2}}}{2 (e^{i b} - e^{i a})})$ $= i π (\frac{e^{- \frac{i b}{2}} - e^{\frac{- i a}{2}}}{(e^{i \frac{a + b}{2}} (e^{i \frac{b - a}{2}} - e^{i - \frac{b - a}{2}})}) = i π \frac{e^{- i b - i \frac{a}{2}} - e^{- i a - i \frac{b}{2}}}{2 i s i n (\frac{a - b}{2})}$
	Commented by mathmax by abdo last updated on 01/Sep/19

	$t h a n k y o u s i r .$
	Commented by mind is power last updated on 01/Sep/19

	$y : r e w e l c o m$
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