Question Number 152795 by mnjuly1970 last updated on 01/Sep/21
$$ \\ $$$$\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({x}\:\right)}{{x}}\left(\frac{{a}^{\:\mathrm{2}} +{cos}^{\:\mathrm{2}} \left({x}\right)}{{b}^{\:\mathrm{2}} +\:{cos}^{\:\mathrm{2}} \left({x}\:\right)}\right){dx}=? \\ $$$$ \\ $$
Answered by Olaf_Thorendsen last updated on 01/Sep/21
![Ο = β«_0 ^β ((sinx)/x)(((a^2 +cos^2 x)/(b^2 +cos^2 x))) dx Let f(x) = ((a^2 +cos^2 x)/(b^2 +cos^2 x)) f is a Οβperiodic function. We can apply the LobatchevskyβDirichlet integral formula : Ο = β«_0 ^β ((sinx)/x)f(x) dx = β«_0 ^(Ο/2) f(x) dx Ο = β«_0 ^(Ο/2) ((a^2 +cos^2 x)/(b^2 +cos^2 x)) dx Ο = β«_0 ^(Ο/2) (1+((a^2 βb^2 )/(b^2 +cos^2 x))) dx Let t = tan(x/2) : Ο = β«_0 ^1 (1+((a^2 βb^2 )/(b^2 +((1βt^2 )/(1+t^2 ))))) ((2dt)/(1+t^2 )) Ο = 2β«_0 ^1 (1+((a^2 βb^2 )/(b^2 +1+(b^2 β1)t^2 ))) dt Let u = (β(β£b^2 β1β£)).t (case bβ 1) : Ο = (2/( (β(β£b^2 β1β£))))β«_0 ^(β(β£b^2 β1β£)) (1+((a^2 βb^2 )/(b^2 +1+u^2 ))) du Ο = (2/( (β(β£b^2 β1β£))))[u+(a^2 βb^2 )(1/( (β(b^2 +1))))arctan((u/( (β(b^2 +1)))))]_0 ^(β(β£b^2 β1β£)) Ο = 2+2((a^2 βb^2 )/( (β(β£b^4 β1β£))))arctan((β(β£((b^2 β1)/(b^2 +1))β£))) do verify my calculous sir](https://www.tinkutara.com/question/Q152798.png)
$$\phi\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}{x}}{{x}}\left(\frac{{a}^{\mathrm{2}} +\mathrm{cos}^{\mathrm{2}} {x}}{{b}^{\mathrm{2}} +\mathrm{cos}^{\mathrm{2}} {x}}\right)\:{dx} \\ $$$$\mathrm{Let}\:{f}\left({x}\right)\:=\:\frac{{a}^{\mathrm{2}} +\mathrm{cos}^{\mathrm{2}} {x}}{{b}^{\mathrm{2}} +\mathrm{cos}^{\mathrm{2}} {x}} \\ $$$${f}\:\mathrm{is}\:\mathrm{a}\:\piβ\mathrm{periodic}\:\mathrm{function}.\:\mathrm{We}\:\mathrm{can} \\ $$$$\mathrm{apply}\:\mathrm{the}\:\mathrm{Lobatchevsky}β\mathrm{Dirichlet} \\ $$$$\mathrm{integral}\:\mathrm{formula}\:: \\ $$$$\phi\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}{x}}{{x}}{f}\left({x}\right)\:{dx}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {f}\left({x}\right)\:{dx} \\ $$$$\phi\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{a}^{\mathrm{2}} +\mathrm{cos}^{\mathrm{2}} {x}}{{b}^{\mathrm{2}} +\mathrm{cos}^{\mathrm{2}} {x}}\:{dx} \\ $$$$\phi\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}+\frac{{a}^{\mathrm{2}} β{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} +\mathrm{cos}^{\mathrm{2}} {x}}\right)\:{dx} \\ $$$$\mathrm{Let}\:{t}\:=\:\mathrm{tan}\frac{{x}}{\mathrm{2}}\:: \\ $$$$\phi\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\frac{{a}^{\mathrm{2}} β{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} +\frac{\mathrm{1}β{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\right)\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\phi\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\frac{{a}^{\mathrm{2}} β{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} +\mathrm{1}+\left({b}^{\mathrm{2}} β\mathrm{1}\right){t}^{\mathrm{2}} }\right)\:{dt} \\ $$$$\mathrm{Let}\:{u}\:=\:\sqrt{\mid{b}^{\mathrm{2}} β\mathrm{1}\mid}.{t}\:\left(\mathrm{case}\:{b}\neq\mathrm{1}\right)\:: \\ $$$$\phi\:=\:\frac{\mathrm{2}}{\:\sqrt{\mid{b}^{\mathrm{2}} β\mathrm{1}\mid}}\int_{\mathrm{0}} ^{\sqrt{\mid{b}^{\mathrm{2}} β\mathrm{1}\mid}} \left(\mathrm{1}+\frac{{a}^{\mathrm{2}} β{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} +\mathrm{1}+{u}^{\mathrm{2}} }\right)\:{du} \\ $$$$\phi\:=\:\frac{\mathrm{2}}{\:\sqrt{\mid{b}^{\mathrm{2}} β\mathrm{1}\mid}}\left[{u}+\left({a}^{\mathrm{2}} β{b}^{\mathrm{2}} \right)\frac{\mathrm{1}}{\:\sqrt{{b}^{\mathrm{2}} +\mathrm{1}}}\mathrm{arctan}\left(\frac{{u}}{\:\sqrt{{b}^{\mathrm{2}} +\mathrm{1}}}\right)\right]_{\mathrm{0}} ^{\sqrt{\mid{b}^{\mathrm{2}} β\mathrm{1}\mid}} \\ $$$$\phi\:=\:\mathrm{2}+\mathrm{2}\frac{{a}^{\mathrm{2}} β{b}^{\mathrm{2}} }{\:\sqrt{\mid{b}^{\mathrm{4}} β\mathrm{1}\mid}}\mathrm{arctan}\left(\sqrt{\mid\frac{{b}^{\mathrm{2}} β\mathrm{1}}{{b}^{\mathrm{2}} +\mathrm{1}}\mid}\right) \\ $$$$\boldsymbol{\mathrm{do}}\:\boldsymbol{\mathrm{verify}}\:\boldsymbol{\mathrm{my}}\:\boldsymbol{\mathrm{calculous}}\:\boldsymbol{\mathrm{sir}} \\ $$
Commented by SANOGO last updated on 01/Sep/21
$${belle}\:{demonstration} \\ $$
Commented by mnjuly1970 last updated on 01/Sep/21
$${bravo}\:{sir}\:{olaf} \\ $$