Question Number 218765 by SdC355 last updated on 15/Apr/25
$$\mathrm{solve} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\:{J}_{\nu} \left({kt}\right){e}^{−{wt}} \mathrm{d}{t}=\mathrm{g}_{\nu,{k}} \left({w}\right) \\ $$
Answered by Nicholas666 last updated on 16/Apr/25
![∫_0 ^∞ J_ν (kt)e^(−wt) dt=g_(ν,k) (w) solution; To solve this integral, we will use the definition of a Bessel function of the first kind and the evaluate the integral; 1. J_ν (x)=Σ_(m=0) ^∞ (((−1)^m )/(m!Γ(m+ν+1)))((x/2))^(2m+ν) 2. ∫_(0 ) ^∞ [Σ_(m=0) ^∞ (((−1)^m )/(m!Γ(m+ν)+1))(((kt)/2))^(2m+ν) ]e^(−wt) dt Σ_(m=0) ^∞ (((−1)^m )/(m!Γ(m+ν+1)))((k/2))^(2m+ν) ∫_(0 ) ^∞ t^(2m+ν) e^(−wt) dt 3. L{t^n }(w)=∫_0 ^∞ t^n e^(−wt) dt=((Γ(n+1))/w^(n+1) ) , with Re(w)>0>−1 ∫_0 ^∞ t^(2m+ν) e^(−wt) dt=((Γ(2m+ν+1))/w^(2m+ν+1) ) 4. g_(ν,k) (w)=Σ_(m=0) ^∞ (((−1)^m )/(m!Γ(m+ν+1)))((k/2))^(2m+ν) ((Γ(2m+ν+1))/w^(2m+ν+1) ) g_(ν,k) (w)=Σ_(m=0) ^∞ (((−1)^m )/(m!))((k/(2w)))^(2m+ν) ((Γ(2m+ν+1))/(Γ(m+ν+1))) (1/w) 5. g_(0,k) (w)=Σ_(m=0) ^∞ (((−1)^m )/(m!))((k/(2w)))^(2m) ((Γ(2m+1))/(Γ(m+1))) (1/w) = ⇒Σ_(m=0) ^∞ (((−1)^m )/(m!))((k/(2w)))^(2m) (((2m)!)/(m!)) (1/w) This doesn′t look like a simple elementary function crane. an alternative approach uses the properties of the Bessel function. There is another way to solve this integral using the integral represantion of the Bessel function; J_ν (x)=(1/π)∫_0 ^π cos(xsinθ−νθ)dθ ∫_0 ^∞ (1/π)∫_0 ^π [cos(ktsinθ−νθ)dθ]e^(−wt) dt (1/π)∫_0 ^π e^(−iνθ) [∫_0 ^∞ cos(ktsinθ)e^(−wt) dt]dθ we know cos(x)=Re(e^(ix) ), ∫_0 ^∞ e^(iktsinθ) e^(−wt) dt=∫_0 ^∞ e^(−(w−iksinθ)t) dt=[(e^(−(w−iksinθ)t) /(−(w−iksinθ)))]_0 ^∞ =(1/(w−iksinθ)) Re((1/(w−iksinθ)))=Re(((w+iksinθ)/(w^2 +k^2 sin^2 θ)))=(w/(w^2 +k^2 sin^2 θ)) so, g_(ν,k) (w)=(1/π)∫_0 ^π ((wcos(νθ))/(w^2 +k^2 sin^2 θ))dθ for the case ν=0 g_(0,k) (w)=(w/π)∫_0 ^π (1/(w^2 +k^2 sin^2 θ))dθ this integral can be solved by substitutation t=tan(θ/2); g_(0,k) (w)=(1/( (√(w^2 +k^2 )))) general results for the of bessel functons of the first kind; g_(ν,k) (w)= (k^ν /( (√(w^2 +k^2 ))(w+(√(w^2 +k^2 )^ν )))) ,with Re(w)>0 , Re(ν)>−1 ∫_(0 ) ^∞ J_ν (kt)e^(−wt) dt = g_(ν,k) (w)(k^ν /((w+(√(w^2 +k^2 )^ν ))(√(w^2 +k^2 )))) ✓](https://www.tinkutara.com/question/Q218863.png)
$$\:\int_{\mathrm{0}} ^{\infty} {J}_{\nu} \left({kt}\right){e}^{−{wt}} {dt}=\mathrm{g}_{\nu,{k}} \left({w}\right) \\ $$$$\:\:{solution}; \\ $$$${To}\:{solve}\:{this}\:{integral},\:{we}\:{will}\:{use}\:{the}\:{definition}\: \\ $$$$\:{of}\:{a}\:{Bessel}\:{function}\:{of}\:{the}\:{first}\:{kind} \\ $$$$\:{and}\:{the}\:{evaluate}\:{the}\:{integral}; \\ $$$$\mathrm{1}.\:\:\:{J}_{\nu} \left({x}\right)=\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} }{{m}!\Gamma\left({m}+\nu+\mathrm{1}\right)}\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}{m}+\nu} \\ $$$$\mathrm{2}.\:\:\int_{\mathrm{0}\:} ^{\infty} \left[\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} }{{m}!\Gamma\left({m}+\nu\right)+\mathrm{1}}\left(\frac{{kt}}{\mathrm{2}}\right)^{\mathrm{2}{m}+\nu} \right]{e}^{−{wt}} {dt}\:\:\:\: \\ $$$$\:\:\:\:\:\:\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} }{{m}!\Gamma\left({m}+\nu+\mathrm{1}\right)}\left(\frac{{k}}{\mathrm{2}}\right)^{\mathrm{2}{m}+\nu} \int_{\mathrm{0}\:} ^{\infty} {t}^{\mathrm{2}{m}+\nu} {e}^{−{wt}} {dt}\:\:\: \\ $$$$\mathrm{3}.\:\:\:\:\mathscr{L}\left\{{t}^{{n}} \right\}\left({w}\right)=\int_{\mathrm{0}} ^{\infty} {t}^{{n}} {e}^{−{wt}} {dt}=\frac{\Gamma\left({n}+\mathrm{1}\right)}{{w}^{{n}+\mathrm{1}} }\:,\:{with}\:{Re}\left({w}\right)>\mathrm{0}>−\mathrm{1}\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} {t}^{\mathrm{2}{m}+\nu} {e}^{−{wt}} {dt}=\frac{\Gamma\left(\mathrm{2}{m}+\nu+\mathrm{1}\right)}{{w}^{\mathrm{2}{m}+\nu+\mathrm{1}} }\: \\ $$$$\mathrm{4}.\:\:\:\:\mathrm{g}_{\nu,{k}} \left({w}\right)=\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} }{{m}!\Gamma\left({m}+\nu+\mathrm{1}\right)}\left(\frac{{k}}{\mathrm{2}}\right)^{\mathrm{2}{m}+\nu} \frac{\Gamma\left(\mathrm{2}{m}+\nu+\mathrm{1}\right)}{{w}^{\mathrm{2}{m}+\nu+\mathrm{1}} }\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\mathrm{g}_{\nu,{k}} \left({w}\right)=\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} }{{m}!}\left(\frac{{k}}{\mathrm{2}{w}}\right)^{\mathrm{2}{m}+\nu} \frac{\Gamma\left(\mathrm{2}{m}+\nu+\mathrm{1}\right)}{\Gamma\left({m}+\nu+\mathrm{1}\right)}\:\frac{\mathrm{1}}{{w}}\:\:\:\: \\ $$$$\mathrm{5}.\:\:\:\mathrm{g}_{\mathrm{0},{k}} \left({w}\right)=\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} }{{m}!}\left(\frac{{k}}{\mathrm{2}{w}}\right)^{\mathrm{2}{m}} \frac{\Gamma\left(\mathrm{2}{m}+\mathrm{1}\right)}{\Gamma\left({m}+\mathrm{1}\right)}\:\frac{\mathrm{1}}{{w}}\:= \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} }{{m}!}\left(\frac{{k}}{\mathrm{2}{w}}\right)^{\mathrm{2}{m}} \frac{\left(\mathrm{2}{m}\right)!}{{m}!}\:\frac{\mathrm{1}}{{w}}\: \\ $$$$\:{This}\:{doesn}'{t}\:{look}\:{like}\:{a}\:{simple}\:{elementary}\:{function}\:{crane}.\:\:\:\:\: \\ $$$$\:\:\:{an}\:{alternative}\:{approach}\:{uses}\:{the}\:{properties}\:{of}\:{the}\:{Bessel}\:{function}.\:\:\:\: \\ $$$${There}\:{is}\:{another}\:{way}\:{to}\:{solve}\: \\ $$$${this}\:{integral}\:{using}\:{the}\:{integral}\: \\ $$$${represantion}\:{of}\:{the}\:{Bessel}\:{function};\:\: \\ $$$$\:\:\:\:{J}_{\nu} \left({x}\right)=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\pi} {cos}\left({xsin}\theta−\nu\theta\right){d}\theta \\ $$$$\:\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\pi} \left[{cos}\left({ktsin}\theta−\nu\theta\right){d}\theta\right]{e}^{−{wt}} {dt} \\ $$$$\:\:\:\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\pi} {e}^{−{i}\nu\theta} \left[\int_{\mathrm{0}} ^{\infty} {cos}\left({ktsin}\theta\right){e}^{−{wt}} {dt}\right]{d}\theta \\ $$$$\:\:\:\:{we}\:{know}\:{cos}\left({x}\right)={Re}\left({e}^{{ix}} \right), \\ $$$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} {e}^{{iktsin}\theta} {e}^{−{wt}} {dt}=\int_{\mathrm{0}} ^{\infty} {e}^{−\left({w}−{iksin}\theta\right){t}} {dt}=\left[\frac{{e}^{−\left({w}−{iksin}\theta\right){t}} }{−\left({w}−{iksin}\theta\right)}\right]_{\mathrm{0}} ^{\infty} =\frac{\mathrm{1}}{{w}−{iksin}\theta}\:\:\:\:\:\: \\ $$$$\:\:\:{Re}\left(\frac{\mathrm{1}}{{w}−{iksin}\theta}\right)={Re}\left(\frac{{w}+{iksin}\theta}{{w}^{\mathrm{2}} +{k}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}\right)=\frac{{w}}{{w}^{\mathrm{2}} +{k}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}\:\:\:\: \\ $$$${so},\:\:\:\:\:\:\:\:\:\mathrm{g}_{\nu,{k}} \left({w}\right)=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\pi} \frac{{wcos}\left(\nu\theta\right)}{{w}^{\mathrm{2}} +{k}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}{d}\theta \\ $$$$\:{for}\:{the}\:{case}\:\nu=\mathrm{0} \\ $$$$\:\:\:\:\mathrm{g}_{\mathrm{0},{k}} \left({w}\right)=\frac{{w}}{\pi}\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{{w}^{\mathrm{2}} +{k}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}{d}\theta \\ $$$$\:\:{this}\:{integral}\:{can}\:{be}\:{solved}\:{by}\:{substitutation}\:{t}={tan}\left(\theta/\mathrm{2}\right); \\ $$$$\:\:\:\:\:\:\:\:\mathrm{g}_{\mathrm{0},{k}} \left({w}\right)=\frac{\mathrm{1}}{\:\sqrt{{w}^{\mathrm{2}} +{k}^{\mathrm{2}} }} \\ $$$$\:\:{general}\:{results}\:{for}\:{the}\:{of}\:{bessel}\:{functons}\:{of}\:{the}\:{first}\:{kind};\:\: \\ $$$$\:\:\mathrm{g}_{\nu,{k}} \left({w}\right)=\:\frac{{k}^{\nu} }{\:\sqrt{{w}^{\mathrm{2}} +{k}^{\mathrm{2}} }\left({w}+\sqrt{\left.{w}^{\mathrm{2}} +{k}^{\mathrm{2}} \right)^{\nu} }\right.}\:,{with}\:{Re}\left({w}\right)>\mathrm{0}\:,\:{Re}\left(\nu\right)>−\mathrm{1}\:\: \\ $$$$ \\ $$$$\:\:\int_{\mathrm{0}\:} ^{\infty} {J}_{\nu} \left({kt}\right){e}^{−{wt}} {dt}\:=\:\mathrm{g}_{\nu,{k}} \left({w}\right)\frac{{k}^{\nu} }{\left({w}+\sqrt{\left.{w}^{\mathrm{2}} +{k}^{\mathrm{2}} \right)^{\nu} }\sqrt{{w}^{\mathrm{2}} +{k}^{\mathrm{2}} }\right.}\:\:\:\checkmark \\ $$$$ \\ $$$$ \\ $$