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

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

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:−{e}^{{ia}} \right)\left({x}^{\mathrm{2}} −{e}^{{ib}} \right)}\:\:{with}\:{a}>\mathrm{0}\:{andb}>\mathrm{0} \\ $$

Commented byMJS last updated on 31/Aug/19

can we calculate ∫_0 ^∞ (dx/((x^2 −α)(x^2 −β))) and then  insert α=e^(ia) ; β=e^(ib) ?

$$\mathrm{can}\:\mathrm{we}\:\mathrm{calculate}\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dx}}{\left({x}^{\mathrm{2}} −\alpha\right)\left({x}^{\mathrm{2}} −\beta\right)}\:\mathrm{and}\:\mathrm{then} \\ $$ $$\mathrm{insert}\:\alpha=\mathrm{e}^{\mathrm{i}{a}} ;\:\beta=\mathrm{e}^{\mathrm{i}{b}} ? \\ $$

Commented bymathmax by abdo last updated on 31/Aug/19

yes sir but you must determine ∫ (dx/(x^2 −z))  with z from C.

$${yes}\:{sir}\:{but}\:{you}\:{must}\:{determine}\:\int\:\frac{{dx}}{{x}^{\mathrm{2}} −{z}}\:\:{with}\:{z}\:{from}\:{C}. \\ $$

Commented bymathmax 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) ))).

$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} −{e}^{{ia}} \right)\left({x}^{\mathrm{2}} −{e}^{{ib}} \right)}\:\Rightarrow\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} −{e}^{{ia}} \right)\left({x}^{\mathrm{2}} −{e}^{{ib}} \right)} \\ $$ $${let}\:\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} −{e}^{{ia}} \right)\left({z}^{\mathrm{2}} −{e}^{{ib}} \right)}\:\Rightarrow\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−{e}^{\frac{{ia}}{\mathrm{2}}} \right)\left({z}+{e}^{\frac{{ia}}{\mathrm{2}}} \right)\left({z}−{e}^{\frac{{ib}}{\mathrm{2}}} \right)\left({z}+{e}^{\frac{{ib}}{\mathrm{2}}} \right)} \\ $$ $${the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}{e}^{\frac{{ia}}{\mathrm{2}}} \:{and}\:\overset{−} {+}{e}^{\frac{{ib}}{\mathrm{2}\:}} \:\:\:{residus}\:{theorem}\:{give} \\ $$ $$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,{e}^{\frac{{ia}}{\mathrm{2}}} \right)+{Res}\left(\varphi,{e}^{\frac{{ib}}{\mathrm{2}}} \right)\right\} \\ $$ $${Res}\left(\varphi,{e}^{\frac{{ia}}{\mathrm{2}}} \right)={lim}_{{z}\rightarrow{e}^{\frac{{ia}}{\mathrm{2}}} } \:\:\left({z}−{e}^{\frac{{ia}}{\mathrm{2}}} \right)\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{e}^{\frac{{ia}}{\mathrm{2}}} \left({e}^{{ia}} −{e}^{{ib}} \right)}\:=\frac{{e}^{−{i}\frac{{a}}{\mathrm{2}}} }{\mathrm{2}\left({e}^{{ia}} −{e}^{{ib}} \right)} \\ $$ $${Res}\left(\varphi,{e}^{\frac{{ib}}{\mathrm{2}}} \right)\:={lim}_{{z}\rightarrow{e}^{\frac{{ib}}{\mathrm{2}}} } \:\:\left({z}−{e}^{\frac{{ib}}{\mathrm{2}}} \right)\varphi\left({z}\right)\:=\frac{{e}^{−\frac{{ib}}{\mathrm{2}}} }{\mathrm{2}\left({e}^{{ib}} \:−{e}^{{ia}} \right)}\:\Rightarrow \\ $$ $$\int_{−\infty} ^{+\infty} \varphi\left(\mathrm{2}\right){dz}\:=\mathrm{2}{i}\pi\left\{\frac{{e}^{−\frac{{ia}}{\mathrm{2}}} }{\mathrm{2}\left({e}^{{ia}} −{e}^{{ib}} \right)}\:+\frac{{e}^{−\frac{{ib}}{\mathrm{2}}} }{\mathrm{2}\left({e}^{{ib}} −{e}^{{ia}} \right)}\right\} \\ $$ $$ \\ $$ $$=\frac{{i}\pi}{{e}^{{ia}} −{e}^{{ib}} }\left\{\:{e}^{−\frac{{ia}}{\mathrm{2}}} −{e}^{−\frac{{ib}}{\mathrm{2}}} \right\}\:=\mathrm{2}{I}\:\Rightarrow \\ $$ $${I}\:=\frac{{i}\pi\left({e}^{−\frac{{ia}}{\mathrm{2}}} −{e}^{−\frac{{ib}}{\mathrm{2}}} \right)}{\mathrm{2}\left({e}^{{ia}} −{e}^{{ib}} \right)}. \\ $$

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))))

$${let}\:\gamma_{{r}} =\left\{{z}\in{C}\:\:{such}\:{that}\mid{z}\mid\leqslant{r}\:\mid{im}\left({z}\right)>\mathrm{0}\right\} \\ $$ $${we}\:{have}\:\int_{\mathrm{0}} ^{+\infty} \frac{{dx}}{\left({x}^{\mathrm{2}} −{e}^{{ia}} \right)\left({x}^{\mathrm{2}} −{e}^{{ib}} \right)}=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \frac{{dx}}{\left({x}^{\mathrm{2}} −{e}^{{ia}} \right)\left({x}^{\mathrm{2}} −{e}^{{ib}} \right)} \\ $$ $${let}\:{f}\left({z}\right)=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} −{e}^{{ia}} \right)\left({z}^{\mathrm{2}} −{e}^{{ib}} \right)} \\ $$ $${f}\left({z}\right)=\frac{\mathrm{1}}{\left({r}^{\mathrm{2}} {e}^{\mathrm{2}{i}\emptyset} −{e}^{{ia}} \right)\left({r}^{\mathrm{2}} {e}^{\mathrm{2}{i}\emptyset} −{e}^{{ib}} \right)}\:\:{over}\:\gamma_{{r}} \\ $$ $$\mid{f}\left({z}\right)\mid\leqslant\frac{\mathrm{1}}{\mid{r}^{\mathrm{2}} −\mathrm{1}\mid^{\mathrm{2}} } \\ $$ $${we}\:{used}\:\mid{z}+{z}'\mid\geqslant\mid\mid{z}\mid−\mid{z}'\mid\mid\:{twice} \\ $$ $$\Rightarrow\int_{\gamma{r}} {f}\left({z}\right){dz}\leqslant\int_{\gamma_{{r}} } \frac{\mathrm{1}}{\left({r}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{dz}=\frac{\pi{r}}{\left({r}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\rightarrow\mathrm{0}={as}\:{r}\rightarrow+\infty \\ $$ $${residu}\:{theorem} \\ $$ $$\left.{poles}\:{of}\:{f}\:{inside}\:\gamma_{{r}} \cup\right]−\infty.+\infty\left[\:{ar}\:{e}^{{i}\frac{{a}}{\mathrm{2}}} .{e}^{\frac{{ib}}{\mathrm{2}}} \right. \\ $$ $${lim}\:\left({z}−{e}^{{i}\frac{{a}}{\mathrm{2}}} \right).{f}\left({z}\right)=\frac{\mathrm{1}}{\mathrm{2}{e}^{{i}\frac{{a}}{\mathrm{2}}} .\left({e}^{{ia}} −{e}^{{ib}} \right)} \\ $$ $${lim}\left({z}−{e}^{{i}\frac{{b}}{\mathrm{2}}} \right){f}\left({z}\right)=\frac{\mathrm{1}}{\mathrm{2}{e}^{{i}\frac{{b}}{\mathrm{2}}} \left({e}^{{ib}} −{e}^{{ia}} \right)} \\ $$ $$\int_{−\infty} ^{+\infty} {f}\left({z}\right){dz}=\mathrm{2}{i}\pi\sum_{{residu}} {f}\left({z}\right)=\mathrm{2}{i}\pi\left(\frac{{e}^{−{i}\frac{{b}}{\mathrm{2}}} }{\left({e}^{{ib}} −{e}^{{ia}} \right)\mathrm{2}}−\frac{{e}^{−\frac{{ia}}{\mathrm{2}}} }{\mathrm{2}\left({e}^{{ib}} −{e}^{{ia}} \right)}\right) \\ $$ $$={i}\pi\left(\frac{{e}^{−\frac{{ib}}{\mathrm{2}}} −{e}^{\frac{−{ia}}{\mathrm{2}}} }{\left({e}^{{i}\frac{{a}+{b}}{\mathrm{2}}} \left({e}^{{i}\frac{{b}−{a}}{\mathrm{2}}} −{e}^{{i}−\frac{{b}−{a}}{\mathrm{2}}} \right)\right.}\right)={i}\pi\frac{{e}^{−{ib}−{i}\frac{{a}}{\mathrm{2}}} −{e}^{−{ia}−{i}\frac{{b}}{\mathrm{2}}} }{\mathrm{2}{isin}\left(\frac{{a}−{b}}{\mathrm{2}}\right)} \\ $$ $$ \\ $$

Commented bymathmax by abdo last updated on 01/Sep/19

thank you sir.

$${thank}\:{you}\:{sir}. \\ $$

Commented bymind is power last updated on 01/Sep/19

y:re welcom

$${y}:{re}\:{welcom} \\ $$ $$ \\ $$

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