Question Number 219060 by MrGaster last updated on 19/Apr/25
$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}^{{m}} }{{x}^{{n}} }{dx},{n}\in\mathbb{N},{m}\in\mathbb{N},{n}\leqslant{m} \\ $$
Answered by Nicholas666 last updated on 19/Apr/25
$$\frac{\pi}{\mathrm{2}} \\ $$
Commented by MrGaster last updated on 19/Apr/25
$$\mathrm{Can}\:\mathrm{you}\:\mathrm{give}\:\mathrm{me}\:\mathrm{a}\:\mathrm{specificl} \\ $$$$\mathrm{caculation}\:\mathrm{process} \\ $$
Answered by MrGaster last updated on 19/Apr/25
=(((2m)!)/s)Π_(k=1) ^n (1/(s^2 −4k^2 )) Proof:I_m =∫_0 ^∞ e^(−sx) sin^m xdx,There is the followingo recursin: I_m =∫_0 ^∞ e^(−sx) sin^m x dx =−(1/s)∫_0 ^∞ sin^m x d(e^(−sx) ) =^(ibp) (m/s)∫_0 ^∞ e^(−sx) sin^(m−1) x cos x dx =−(m/s^2 )∫_0 ^∞ sin^(m−1) x cos x d(e^(−sx) ) =^(idp) (m/s^2 )∫_0 ^∞ e^(−sx) ((m−1)sin^(m−2) x cos^2 x−sin^m x)dx =((m(m−1))/s^2 )I_(m−2) −(m^2 /s^2 )I_m ⇒I_(2m) =((2m(2m−1))/((2m)^2 +s^2 ))I_(m−2) =…=I_0 (2m)!Π_(k=1) ^m (1/(s^2 +4k^2 ))=(((2m)!)/s)Π_(k=1) ^n (1/(s^2 −4k^2 )) Lemmma 2: L[sin^(2m−1) x](s)=(2m−1)!Π_(k=1) ^m (1/(s^2 +(2k−1)^2 )) L[sin^(2m−1) x](s)=(2m−1)!Π_(k=1) ^m (1/(s^2 +(2k−1)^2 )) Proof: I_(2m−1) =I_1 (2m−1)!Π_(k=2) ^m (1/(s^2 +(2k+1)))=(2m−1)!Π_(k=1) ^m (1/(s^2 +(2k+1)^2 )) Lemma 3:L^(−1) [x^(−n) ](s)=(s^(n+1) /(Γ(n))) Proof:L[x^(s−1) ](s)=∫_0 ^∞ e^(−sx) z^(n−1) dx=^(sx x) s^(−n) ∫_0 ^∞ x^(n−1) e^(−x) =s^(−n) Γ(n) Lemma 4(Laplace Cross product formula) ∫_0 ^∞ f(x)g(x)dx=∫_0 ^∞ L[f(x)]L^(−1) [g(x)]ds ∫_0 ^∞ f(x)g(x)dx=∫_0 ^∞ L[f(x)]L^(−1) [g(x)]ds ∫_0 ^∞ f(x)L[h(s)](x)dx=∫_0 ^∞ f(x)∫_0 ^∞ e^(−sx) h(s)dsdx =∫_0 ^∞ h(s)∫_0 ^∞ e^(−sx) f(x)dxds=∫_0 ^∞ h(s)L[f(x)](s)ds Let h(s)=L^(−1) [g(x)],Lemma 4 is correct. Now may wish to analyze thesituation: 1.m,nis an even number so m 2m,n 2n Lemma 1&Lemma 3&Lemma 4 ⇒∫_0 ^∞ ((sin^(2m) x)/x^(2n) )dx=∫_0 ^∞ L[sin^(2m) x]L^(−1) [x^(−2n) ]dx=(((2m)!)/((2n−1)!))∫_0 ^∞ Π_(k=1) ^n (1/((s^2 +4k^2 )))∙s^(2n−2) ds Let f(z)=Π_(k=1) ^m (1/((z^2 +4k^2 )))∙z^(2m−2) ∧∮_C f(z)dz=(∫_C_R +∫_(−∞) ^(+∞) )f(z)dz,C_R ={z:∣z∣=R,J(z)>0} ∵Jordan lemma ∫_C_R f(z)dz=0 ∴∫_(−∞) ^(+∞) f(s)ds=2πiΣ_(j=1) ^m Res[f(z),2j∙i]=2πiΣ_(j=1) ^m ((((2j∙i)^(2m−2) )/(4j∙i))Π_(k≠j) (1/(4k^2 −4j^2 ))) =2π(−1)^(n−m−1) Σ_(j=1) ^m j^(2n−3) Π_(k≠j) (1/(k^2 ∙j^2 ))=π(−1)^n 4^(n−m) Σ_(j=1) ^m (((−1)^j j^(2n−1) )/((m+j)!(m−j)!)) Substitute it into the formulaat the beginning, andthere is: ∫_0 ^∞ ((sin^(2m) x)/x^(2n) )dx=π(−1)^n ((4^(n−m) (2m)!)/(2(2n−1)!))Σ_(j=1) ^m (((−1)^j j^(2n−1) )/((m−j)!(m+j)!)) 2.m,n∈odd∧m 2m−1,n 2n−1 ∵Lemma 1&Lemma 3&Lemma 4 ⇒∫_0 ^∞ ((sin^(2m−1) x)/x^(2n−1) )dx=∫_0 ^∞ L[sin^(2m−1) x]L^(−1) [x^(1−2n) ]dx =(((2m−1)!)/((2n−2)!))∫_0 ^∞ Π_(k=1) ^m (1/(s^2 +(2k−1)^2 ))∙s^(2n−2) ds Let g(z)=Π_(k=1) ^m (1/(z^2 +(2k−1)^2 ))∙z^(2n−2) ⇒∮_C g(z)dz=(∫_C +∫_(−∞) ^(+∞) )g(z)dz,C_R ={z:∣z∣=R,J(z)>0} ∵Jordan lemma ∫_C_R f(z)dz=0 ⇒∫_(−∞) ^(+∞) g(s)ds=2πiΣ_(j=1) ^m Res[g(z),(2j−11()i] =2πiΣ_(j=1) ^m ((([(2j−1)i]^(2n−2) )/(2(2j−1)i))Π_(k≠j) (1/((2k−1)^2 −(2j−1)^2 ))) =π(−1)^(n−1) 4^(1−m) Σ_(j=1) ^m (2j−1)^(2n−3) Π_(k≠j) (1/((k−j)(k+j−1))) =π(−1)^n 4^(1−m) Σ_(j=1) ^m (((−1)^j (2j−1)^(2m−2) )/((m+j−1)(m−j)!)) Substitute it into the formulaat the beginning andthere is: ∫_0 ^∞ ((sin^(2m−1) )/x^(2m−1) )dx=π(−1)((4^(1−m) (2m−1)!)/(2(2n−2)!))Σ_(j=1) ^m (((−1)^j (2j−1)^(2n−2) )/((m−j)!(m+j−1)!)) 3.m∈even∧n∈odd,m 2m,n 2n+1 ∫_0 ^∞ ((sin^(2m) x)/x^(2n−1) )dx=∫_0 ^∞ L[sin^(2n) x]L^(−1) [x^(−2n−1) ]dx =(((2m)!)/((2n)!))∫_0 ^∞ Π_(k=1) ^n (1/((s^2 +4k^2 )))∙s^(2n−1) ds=^(s^2 s) (((2m)!)/(2(2n)!))∫_0 ^∞ Π_(k=1) ^m (1/(s+4k^2 ))∙s^(n−1) ds By partial fractionalo decompsition of thed integran function there is: ∫_0 ^∞ Π_(k=1) ^m (1/(s+4k^2 ))∙s^(n−1) ds =(−1)^(n−1) ∫_0 ^∞ Σ_(j=1) ^m (Π_(k≠j) (1/(4k^2 −4j^2 )) (((2j)^(2n−2) )/(s+4j^2 )))ds =(−1)^(n−1) 4^(n−m) Σ_(j=1) ^m Π_(k≠j) (1/(k^2 −j^2 ))∙j^(2n−2) log(s+4j^2 )∣_0 ^∞ =(−1)^n 4^(n−m+1) Σ_(j=1) ^m (((−1)^j j^(2n) )/((m+j)!(m−j)!))∙log 2j =(−1)^n 4^(n−m+1) Σ_(j=1) ^m (((−1)^j j^(2n) )/((m+j)!(m−j)))∙log j ∴∫((sin^(2m) )/x^(2n−1) )dx=(−1)^n ((4^(n−m+1) (2m)!)/(2(2n)!))Σ_(j=1) ^m (((−1)^j j^(2n) )/((m+j)!(m−j)))∙log j 4m∈odd,n∈even,m 2m−1,n 2n ∫_0 ^∞ ((sin^(2m) x)/x^(2n−1) )dx=∫_0 ^∞ L[sin^(2m−1) x]L^(−1) [x^(−2n) ]dx =(((2m−1)!)/((2n−1)!))∫_0 ^∞ Π_(k=1) ^m (1/(s^2 +(2k−1)^2 ))∙s^(2n−1) ds =^(s^2 s) (((2m−1)!)/(2(2n−1)!))∫_0 ^∞ Π_(k=1) ^n (1/(s+(2k−1)^2 ))∙s^(n−1) ds By partial fractionalo decompsition of thed integran function there is: ∫_0 ^∞ Π_(k=1) ^m (1/(s+(2k−1)^2 ))∙s^(n−1) ds =(−1)^(n−1) ∫_0 ^∞ Σ_(j=1) ^m Π_(k≠j) (1/((k−j)(k+j−1)))∙(2j−1)^(2n−2) log(s+(2j+1)^2 )∣_0 ^∞ =(−1)^n 4^(1−m) Σ_(j=1) ^m (((−1)^j (2j−1)^(2n−1) )/((m+j−1)!(m−j)))∙log(2j−1) Substitute it into the formulaat the beginning andthere is: ∫_0 ^∞ ((sin^(2m−1) x)/x^(2n) )dx=(−1)^n ((4^(1−m) (2m−1)!)/(2(2n−1)!))Σ_(j=1) ^m (((−1)^j (2j−1)^(2n−1) log(2j−1))/((m+j−1)!(m−j)!)) Through the above analysise w get the final result: ∫_0 ^∞ ((sin^(2m) x)/x^(2n) )dx=((4^(n−m) (2m)!π)/(2(2n−1)!))Σ_(j=1) ^∞ (((−1)^(n−j) j^(2n−1) )/((m−j)!(m+j)!)) ∫_0 ^∞ ((sin^(2m−1) x)/x^(2n−1) )dx=((4^(1−m) (2m−1)!π)/(2(2n−2)!))Σ_(j=1) ^m (((−1)^(n−j) (2j−1)^(2n−2) )/((m−j)!(m+j−1)!)) ∫_0 ^∞ ((sin^(2m) x)/x^(2n−1) )dx=((4^(n−m+1) (2m)!)/(2(2n)!))Σ_(j=1) ^m (((−1)^(n−j) j^(2n) log j)/((m+j)!(m−j)!)) ∫_0 ^∞ ((sin^(2m−1) x)/x^(2n) )dx=((4^(1−m) (2m−1)!)/(2(2n−1)!))Σ_(j=1) ^m (((−1)^(n−j) (2j−1)^(2n−1) log(2j−1))/((m+j−1)!(m−j)!))](https://www.tinkutara.com/question/Q219073.png)
$$\mathrm{Lemma}\:\mathrm{1}: \\ $$$$\mathscr{L}\left[\mathrm{sin}^{\mathrm{2}{m}} {x}\right]\left({s}\right)=\frac{\left(\mathrm{2}{m}\right)!}{{s}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{\mathrm{1}}{{s}^{\mathrm{2}} −\mathrm{4}{k}^{\mathrm{2}} } \\ $$$$\mathrm{Proof}:{I}_{{m}} =\int_{\mathrm{0}} ^{\infty} {e}^{−{sx}} \mathrm{sin}^{{m}} {xdx},\mathrm{There}\:\mathrm{is}\:\mathrm{the}\:\mathrm{followingo} \\ $$$$\mathrm{recursin}: \\ $$$${I}_{{m}} =\int_{\mathrm{0}} ^{\infty} {e}^{−{sx}} \mathrm{sin}^{{m}} {x}\:{dx} \\ $$$$=−\frac{\mathrm{1}}{{s}}\int_{\mathrm{0}} ^{\infty} \mathrm{sin}^{{m}} {x}\:{d}\left({e}^{−{sx}} \right) \\ $$$$\overset{{ibp}} {=}\frac{{m}}{{s}}\int_{\mathrm{0}} ^{\infty} {e}^{−{sx}} \mathrm{sin}^{{m}−\mathrm{1}} {x}\:\mathrm{cos}\:{x}\:{dx} \\ $$$$=−\frac{{m}}{{s}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \mathrm{sin}^{{m}−\mathrm{1}} {x}\:\mathrm{cos}\:{x}\:{d}\left({e}^{−{sx}} \right) \\ $$$$\overset{{idp}} {=}\frac{{m}}{{s}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} {e}^{−{sx}} \left(\left({m}−\mathrm{1}\right)\mathrm{sin}^{{m}−\mathrm{2}} {x}\:\mathrm{cos}^{\mathrm{2}} {x}−\mathrm{sin}^{{m}} {x}\right){dx} \\ $$$$=\frac{{m}\left({m}−\mathrm{1}\right)}{{s}^{\mathrm{2}} }{I}_{{m}−\mathrm{2}} −\frac{{m}^{\mathrm{2}} }{{s}^{\mathrm{2}} }{I}_{{m}} \\ $$$$\Rightarrow{I}_{\mathrm{2}{m}} =\frac{\mathrm{2}{m}\left(\mathrm{2}{m}−\mathrm{1}\right)}{\left(\mathrm{2}{m}\right)^{\mathrm{2}} +{s}^{\mathrm{2}} }{I}_{{m}−\mathrm{2}} =\ldots={I}_{\mathrm{0}} \left(\mathrm{2}{m}\right)!\underset{{k}=\mathrm{1}} {\overset{{m}} {\prod}}\frac{\mathrm{1}}{{s}^{\mathrm{2}} +\mathrm{4}{k}^{\mathrm{2}} }=\frac{\left(\mathrm{2}{m}\right)!}{{s}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{\mathrm{1}}{{s}^{\mathrm{2}} −\mathrm{4}{k}^{\mathrm{2}} } \\ $$$$\mathrm{Lemmma}\:\mathrm{2}: \\ $$$$\mathscr{L}\left[\mathrm{sin}^{\mathrm{2}{m}−\mathrm{1}} {x}\right]\left({s}\right)=\left(\mathrm{2}{m}−\mathrm{1}\right)!\underset{{k}=\mathrm{1}} {\overset{{m}} {\prod}}\frac{\mathrm{1}}{{s}^{\mathrm{2}} +\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathscr{L}\left[\mathrm{sin}^{\mathrm{2}{m}−\mathrm{1}} {x}\right]\left({s}\right)=\left(\mathrm{2}{m}−\mathrm{1}\right)!\underset{{k}=\mathrm{1}} {\overset{{m}} {\prod}}\frac{\mathrm{1}}{{s}^{\mathrm{2}} +\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{Proof}: \\ $$$${I}_{\mathrm{2}{m}−\mathrm{1}} ={I}_{\mathrm{1}} \left(\mathrm{2}{m}−\mathrm{1}\right)!\underset{{k}=\mathrm{2}} {\overset{{m}} {\prod}}\frac{\mathrm{1}}{{s}^{\mathrm{2}} +\left(\mathrm{2}{k}+\mathrm{1}\right)}=\left(\mathrm{2}{m}−\mathrm{1}\right)!\underset{{k}=\mathrm{1}} {\overset{{m}} {\prod}}\frac{\mathrm{1}}{{s}^{\mathrm{2}} +\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{Lemma}\:\mathrm{3}:\mathscr{L}^{−\mathrm{1}} \left[{x}^{−{n}} \right]\left({s}\right)=\frac{{s}^{{n}+\mathrm{1}} }{\Gamma\left({n}\right)} \\ $$$$\mathrm{Proof}:\mathscr{L}\left[{x}^{{s}−\mathrm{1}} \right]\left({s}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{−{sx}} {z}^{{n}−\mathrm{1}} {dx}\overset{{sx} {x}} {=}{s}^{−{n}} \int_{\mathrm{0}} ^{\infty} {x}^{{n}−\mathrm{1}} {e}^{−{x}} ={s}^{−{n}} \Gamma\left({n}\right) \\ $$$$\mathrm{Lemma}\:\mathrm{4}\left(\mathrm{Laplace}\:\mathrm{Cross}\:\mathrm{product}\:\mathrm{formula}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} {f}\left({x}\right){g}\left({x}\right){dx}=\int_{\mathrm{0}} ^{\infty} \mathscr{L}\left[{f}\left({x}\right)\right]\mathscr{L}^{−\mathrm{1}} \left[{g}\left({x}\right)\right]{ds} \\ $$$$\int_{\mathrm{0}} ^{\infty} {f}\left({x}\right){g}\left({x}\right){dx}=\int_{\mathrm{0}} ^{\infty} \mathscr{L}\left[{f}\left({x}\right)\right]\mathscr{L}^{−\mathrm{1}} \left[{g}\left({x}\right)\right]{ds} \\ $$$$\int_{\mathrm{0}} ^{\infty} {f}\left({x}\right)\mathscr{L}\left[{h}\left({s}\right)\right]\left({x}\right){dx}=\int_{\mathrm{0}} ^{\infty} {f}\left({x}\right)\int_{\mathrm{0}} ^{\infty} {e}^{−{sx}} {h}\left({s}\right){dsdx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} {h}\left({s}\right)\int_{\mathrm{0}} ^{\infty} {e}^{−{sx}} {f}\left({x}\right){dxds}=\int_{\mathrm{0}} ^{\infty} {h}\left({s}\right)\mathscr{L}\left[{f}\left({x}\right)\right]\left({s}\right){ds} \\ $$$$\mathrm{Let}\:{h}\left({s}\right)=\mathscr{L}^{−\mathrm{1}} \left[{g}\left({x}\right)\right],\mathrm{Lemma}\:\mathrm{4}\:\mathrm{is}\:\mathrm{correct}. \\ $$$$\mathrm{Now}\:\mathrm{may}\:\mathrm{wish}\:\mathrm{to}\:\mathrm{analyze}\: \\ $$$$\mathrm{thesituation}: \\ $$$$\mathrm{1}.{m},{n}\mathrm{is}\:\mathrm{an}\:\mathrm{even}\:\mathrm{number}\:\mathrm{so}\:{m} \mathrm{2}{m},{n} \mathrm{2}{n} \\ $$$$\mathrm{Lemma}\:\mathrm{1\&Lemma}\:\mathrm{3\&Lemma}\:\mathrm{4} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}^{\mathrm{2}{m}} {x}}{{x}^{\mathrm{2}{n}} }{dx}=\int_{\mathrm{0}} ^{\infty} \mathscr{L}\left[\mathrm{sin}^{\mathrm{2}{m}} {x}\right]\mathscr{L}^{−\mathrm{1}} \left[{x}^{−\mathrm{2}{n}} \right]{dx}=\frac{\left(\mathrm{2}{m}\right)!}{\left(\mathrm{2}{n}−\mathrm{1}\right)!}\int_{\mathrm{0}} ^{\infty} \underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{\mathrm{1}}{\left({s}^{\mathrm{2}} +\mathrm{4}{k}^{\mathrm{2}} \right)}\centerdot{s}^{\mathrm{2}{n}−\mathrm{2}} {ds} \\ $$$$\mathrm{Let}\:{f}\left({z}\right)=\underset{{k}=\mathrm{1}} {\overset{{m}} {\prod}}\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} +\mathrm{4}{k}^{\mathrm{2}} \right)}\centerdot{z}^{\mathrm{2}{m}−\mathrm{2}} \wedge\oint_{{C}} {f}\left({z}\right){dz}=\left(\int_{{C}_{{R}} } +\int_{−\infty} ^{+\infty} \right){f}\left({z}\right){dz},{C}_{{R}} =\left\{{z}:\mid{z}\mid={R},\mathfrak{J}\left({z}\right)>\mathrm{0}\right\} \\ $$$$\because\mathrm{Jordan}\:\mathrm{lemma}\: \int_{{C}_{{R}} } {f}\left({z}\right){dz}=\mathrm{0} \\ $$$$\therefore\int_{−\infty} ^{+\infty} {f}\left({s}\right){ds}=\mathrm{2}\pi{i}\underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}\mathrm{Res}\left[{f}\left({z}\right),\mathrm{2}{j}\centerdot{i}\right]=\mathrm{2}\pi{i}\underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}\left(\frac{\left(\mathrm{2}{j}\centerdot{i}\right)^{\mathrm{2}{m}−\mathrm{2}} }{\mathrm{4}{j}\centerdot{i}}\underset{{k}\neq{j}} {\prod}\frac{\mathrm{1}}{\mathrm{4}{k}^{\mathrm{2}} −\mathrm{4}{j}^{\mathrm{2}} }\right) \\ $$$$=\mathrm{2}\pi\left(−\mathrm{1}\right)^{{n}−{m}−\mathrm{1}} \underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}{j}^{\mathrm{2}{n}−\mathrm{3}} \underset{{k}\neq{j}} {\prod}\frac{\mathrm{1}}{{k}^{\mathrm{2}} \centerdot{j}^{\mathrm{2}} }=\pi\left(−\mathrm{1}\right)^{{n}} \mathrm{4}^{{n}−{m}} \underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{j}} {j}^{\mathrm{2}{n}−\mathrm{1}} }{\left({m}+{j}\right)!\left({m}−{j}\right)!} \\ $$$$\mathrm{Substitute}\:\mathrm{it}\:\mathrm{into}\:\mathrm{the}\: \\ $$$$\mathrm{formulaat}\:\mathrm{the}\:\mathrm{beginning},\: \\ $$$$\mathrm{andthere}\:\mathrm{is}: \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}^{\mathrm{2}{m}} {x}}{{x}^{\mathrm{2}{n}} }{dx}=\pi\left(−\mathrm{1}\right)^{{n}} \frac{\mathrm{4}^{{n}−{m}} \left(\mathrm{2}{m}\right)!}{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)!}\underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{j}} {j}^{\mathrm{2}{n}−\mathrm{1}} }{\left({m}−{j}\right)!\left({m}+{j}\right)!} \\ $$$$\mathrm{2}.{m},{n}\in\mathrm{odd}\wedge{m} \mathrm{2}{m}−\mathrm{1},{n} \mathrm{2}{n}−\mathrm{1} \\ $$$$\because\mathrm{Lemma}\:\mathrm{1\&Lemma}\:\mathrm{3\&Lemma}\:\mathrm{4} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}^{\mathrm{2}{m}−\mathrm{1}} {x}}{{x}^{\mathrm{2}{n}−\mathrm{1}} }{dx}=\int_{\mathrm{0}} ^{\infty} \mathscr{L}\left[\mathrm{sin}^{\mathrm{2}{m}−\mathrm{1}} {x}\right]\mathscr{L}^{−\mathrm{1}} \left[{x}^{\mathrm{1}−\mathrm{2}{n}} \right]{dx} \\ $$$$=\frac{\left(\mathrm{2}{m}−\mathrm{1}\right)!}{\left(\mathrm{2}{n}−\mathrm{2}\right)!}\int_{\mathrm{0}} ^{\infty} \underset{{k}=\mathrm{1}} {\overset{{m}} {\prod}}\frac{\mathrm{1}}{{s}^{\mathrm{2}} +\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{2}} }\centerdot{s}^{\mathrm{2}{n}−\mathrm{2}} {ds} \\ $$$$\mathrm{Let}\:{g}\left({z}\right)=\underset{{k}=\mathrm{1}} {\overset{{m}} {\prod}}\frac{\mathrm{1}}{{z}^{\mathrm{2}} +\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{2}} }\centerdot{z}^{\mathrm{2}{n}−\mathrm{2}} \\ $$$$\Rightarrow\oint_{{C}} {g}\left({z}\right){dz}=\left(\int_{{C}} +\int_{−\infty} ^{+\infty} \right){g}\left({z}\right){dz},{C}_{{R}} =\left\{{z}:\mid{z}\mid={R},\mathfrak{J}\left({z}\right)>\mathrm{0}\right\} \\ $$$$\because\mathrm{Jordan}\:\mathrm{lemma} \int_{{C}_{{R}} } {f}\left({z}\right){dz}=\mathrm{0} \\ $$$$\Rightarrow\int_{−\infty} ^{+\infty} {g}\left({s}\right){ds}=\mathrm{2}\pi{i}\underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}\mathrm{Res}\left[{g}\left({z}\right),\left(\mathrm{2}{j}−\mathrm{11}\left(\right){i}\right]\right. \\ $$$$=\mathrm{2}\pi{i}\underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}\left(\frac{\left[\left(\mathrm{2}{j}−\mathrm{1}\right){i}\right]^{\mathrm{2}{n}−\mathrm{2}} }{\mathrm{2}\left(\mathrm{2}{j}−\mathrm{1}\right){i}}\underset{{k}\neq{j}} {\prod}\frac{\mathrm{1}}{\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{2}{j}−\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$$=\pi\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \mathrm{4}^{\mathrm{1}−{m}} \underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}\left(\mathrm{2}{j}−\mathrm{1}\right)^{\mathrm{2}{n}−\mathrm{3}} \underset{{k}\neq{j}} {\prod}\frac{\mathrm{1}}{\left({k}−{j}\right)\left({k}+{j}−\mathrm{1}\right)} \\ $$$$=\pi\left(−\mathrm{1}\right)^{{n}} \mathrm{4}^{\mathrm{1}−{m}} \underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{j}} \left(\mathrm{2}{j}−\mathrm{1}\right)^{\mathrm{2}{m}−\mathrm{2}} }{\left({m}+{j}−\mathrm{1}\right)\left({m}−{j}\right)!} \\ $$$$\mathrm{Substitute}\:\mathrm{it}\:\mathrm{into}\:\mathrm{the}\: \\ $$$$\mathrm{formulaat}\:\mathrm{the}\:\mathrm{beginning}\: \\ $$$$\mathrm{andthere}\:\mathrm{is}: \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}^{\mathrm{2}{m}−\mathrm{1}} }{{x}^{\mathrm{2}{m}−\mathrm{1}} }{dx}=\pi\left(−\mathrm{1}\right)\frac{\mathrm{4}^{\mathrm{1}−{m}} \left(\mathrm{2}{m}−\mathrm{1}\right)!}{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{2}\right)!}\underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{j}} \left(\mathrm{2}{j}−\mathrm{1}\right)^{\mathrm{2}{n}−\mathrm{2}} }{\left({m}−{j}\right)!\left({m}+{j}−\mathrm{1}\right)!} \\ $$$$\mathrm{3}.{m}\in\mathrm{even}\wedge{n}\in\mathrm{odd},{m} \mathrm{2}{m},{n} \mathrm{2}{n}+\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}^{\mathrm{2}{m}} {x}}{{x}^{\mathrm{2}{n}−\mathrm{1}} }{dx}=\int_{\mathrm{0}} ^{\infty} \mathscr{L}\left[\mathrm{sin}^{\mathrm{2}{n}} {x}\right]\mathscr{L}^{−\mathrm{1}} \left[{x}^{−\mathrm{2}{n}−\mathrm{1}} \right]{dx} \\ $$$$=\frac{\left(\mathrm{2}{m}\right)!}{\left(\mathrm{2}{n}\right)!}\int_{\mathrm{0}} ^{\infty} \underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{\mathrm{1}}{\left({s}^{\mathrm{2}} +\mathrm{4}{k}^{\mathrm{2}} \right)}\centerdot{s}^{\mathrm{2}{n}−\mathrm{1}} {ds}\overset{{s}^{\mathrm{2}} {s}} {=}\frac{\left(\mathrm{2}{m}\right)!}{\mathrm{2}\left(\mathrm{2}{n}\right)!}\int_{\mathrm{0}} ^{\infty} \underset{{k}=\mathrm{1}} {\overset{{m}} {\prod}}\frac{\mathrm{1}}{{s}+\mathrm{4}{k}^{\mathrm{2}} }\centerdot{s}^{{n}−\mathrm{1}} {ds} \\ $$$$\mathrm{By}\:\mathrm{partial}\:\mathrm{fractionalo} \\ $$$$\mathrm{decompsition}\:\mathrm{of}\:\mathrm{thed} \\ $$$$\mathrm{integran}\:\mathrm{function}\:\mathrm{there}\:\mathrm{is}: \\ $$$$\int_{\mathrm{0}} ^{\infty} \underset{{k}=\mathrm{1}} {\overset{{m}} {\prod}}\frac{\mathrm{1}}{{s}+\mathrm{4}{k}^{\mathrm{2}} }\centerdot{s}^{{n}−\mathrm{1}} {ds} \\ $$$$=\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \int_{\mathrm{0}} ^{\infty} \underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}\left(\underset{{k}\neq{j}} {\prod}\frac{\mathrm{1}}{\mathrm{4}{k}^{\mathrm{2}} −\mathrm{4}{j}^{\mathrm{2}} }\:\frac{\left(\mathrm{2}{j}\right)^{\mathrm{2}{n}−\mathrm{2}} }{{s}+\mathrm{4}{j}^{\mathrm{2}} }\right){ds} \\ $$$$=\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \mathrm{4}^{{n}−{m}} \underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}\underset{{k}\neq{j}} {\prod}\frac{\mathrm{1}}{{k}^{\mathrm{2}} −{j}^{\mathrm{2}} }\centerdot{j}^{\mathrm{2}{n}−\mathrm{2}} \mathrm{log}\left({s}+\mathrm{4}{j}^{\mathrm{2}} \right)\mid_{\mathrm{0}} ^{\infty} \\ $$$$=\left(−\mathrm{1}\right)^{{n}} \mathrm{4}^{{n}−{m}+\mathrm{1}} \underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{j}} {j}^{\mathrm{2}{n}} }{\left({m}+{j}\right)!\left({m}−{j}\right)!}\centerdot\mathrm{log}\:\mathrm{2}{j} \\ $$$$=\left(−\mathrm{1}\right)^{{n}} \mathrm{4}^{{n}−{m}+\mathrm{1}} \underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{j}} {j}^{\mathrm{2}{n}} }{\left({m}+{j}\right)!\left({m}−{j}\right)}\centerdot\mathrm{log}\:{j} \\ $$$$\therefore\int\frac{\mathrm{sin}^{\mathrm{2}{m}} }{{x}^{\mathrm{2}{n}−\mathrm{1}} }{dx}=\left(−\mathrm{1}\right)^{{n}} \frac{\mathrm{4}^{{n}−{m}+\mathrm{1}} \left(\mathrm{2}{m}\right)!}{\mathrm{2}\left(\mathrm{2}{n}\right)!}\underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{j}} {j}^{\mathrm{2}{n}} }{\left({m}+{j}\right)!\left({m}−{j}\right)}\centerdot\mathrm{log}\:{j} \\ $$$$\mathrm{4}{m}\in\mathrm{odd},{n}\in\mathrm{even},{m} \mathrm{2}{m}−\mathrm{1},{n} \mathrm{2}{n} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}^{\mathrm{2}{m}} {x}}{{x}^{\mathrm{2}{n}−\mathrm{1}} }{dx}=\int_{\mathrm{0}} ^{\infty} \mathscr{L}\left[\mathrm{sin}^{\mathrm{2}{m}−\mathrm{1}} {x}\right]\mathscr{L}^{−\mathrm{1}} \left[{x}^{−\mathrm{2}{n}} \right]{dx} \\ $$$$=\frac{\left(\mathrm{2}{m}−\mathrm{1}\right)!}{\left(\mathrm{2}{n}−\mathrm{1}\right)!}\int_{\mathrm{0}} ^{\infty} \underset{{k}=\mathrm{1}} {\overset{{m}} {\prod}}\frac{\mathrm{1}}{{s}^{\mathrm{2}} +\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{2}} }\centerdot{s}^{\mathrm{2}{n}−\mathrm{1}} {ds} \\ $$$$\overset{{s}^{\mathrm{2}} \: {s}} {=}\frac{\left(\mathrm{2}{m}−\mathrm{1}\right)!}{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)!}\int_{\mathrm{0}} ^{\infty} \underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{\mathrm{1}}{{s}+\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{2}} }\centerdot{s}^{{n}−\mathrm{1}} {ds} \\ $$$$\mathrm{By}\:\mathrm{partial}\:\mathrm{fractionalo} \\ $$$$\mathrm{decompsition}\:\mathrm{of}\:\mathrm{thed} \\ $$$$\mathrm{integran}\:\mathrm{function}\:\mathrm{there}\:\mathrm{is}: \\ $$$$\int_{\mathrm{0}} ^{\infty} \underset{{k}=\mathrm{1}} {\overset{{m}} {\prod}}\frac{\mathrm{1}}{{s}+\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{2}} }\centerdot{s}^{{n}−\mathrm{1}} {ds} \\ $$$$=\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \int_{\mathrm{0}} ^{\infty} \underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}\underset{{k}\neq{j}} {\prod}\frac{\mathrm{1}}{\left({k}−{j}\right)\left({k}+{j}−\mathrm{1}\right)}\centerdot\left(\mathrm{2}{j}−\mathrm{1}\right)^{\mathrm{2}{n}−\mathrm{2}} \mathrm{log}\left({s}+\left(\mathrm{2}{j}+\mathrm{1}\right)^{\mathrm{2}} \right)\mid_{\mathrm{0}} ^{\infty} \\ $$$$=\left(−\mathrm{1}\right)^{{n}} \mathrm{4}^{\mathrm{1}−{m}} \underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{j}} \left(\mathrm{2}{j}−\mathrm{1}\right)^{\mathrm{2}{n}−\mathrm{1}} }{\left({m}+{j}−\mathrm{1}\right)!\left({m}−{j}\right)}\centerdot\mathrm{log}\left(\mathrm{2}{j}−\mathrm{1}\right) \\ $$$$\mathrm{Substitute}\:\mathrm{it}\:\mathrm{into}\:\mathrm{the}\: \\ $$$$\mathrm{formulaat}\:\mathrm{the}\:\mathrm{beginning}\: \\ $$$$\mathrm{andthere}\:\mathrm{is}: \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}^{\mathrm{2}{m}−\mathrm{1}} {x}}{{x}^{\mathrm{2}{n}} }{dx}=\left(−\mathrm{1}\right)^{{n}} \frac{\mathrm{4}^{\mathrm{1}−{m}} \left(\mathrm{2}{m}−\mathrm{1}\right)!}{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)!}\underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{j}} \left(\mathrm{2}{j}−\mathrm{1}\right)^{\mathrm{2}{n}−\mathrm{1}} \mathrm{log}\left(\mathrm{2}{j}−\mathrm{1}\right)}{\left({m}+{j}−\mathrm{1}\right)!\left({m}−{j}\right)!} \\ $$$$\mathrm{Through}\:\mathrm{the}\:\mathrm{above}\:\mathrm{analysise} \\ $$$$\mathrm{w}\:\mathrm{get}\:\mathrm{the}\:\mathrm{final}\:\mathrm{result}: \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}^{\mathrm{2}{m}} {x}}{{x}^{\mathrm{2}{n}} }{dx}=\frac{\mathrm{4}^{{n}−{m}} \left(\mathrm{2}{m}\right)!\pi}{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)!}\underset{{j}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−{j}} {j}^{\mathrm{2}{n}−\mathrm{1}} }{\left({m}−{j}\right)!\left({m}+{j}\right)!} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}^{\mathrm{2}{m}−\mathrm{1}} {x}}{{x}^{\mathrm{2}{n}−\mathrm{1}} }{dx}=\frac{\mathrm{4}^{\mathrm{1}−{m}} \left(\mathrm{2}{m}−\mathrm{1}\right)!\pi}{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{2}\right)!}\underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−{j}} \left(\mathrm{2}{j}−\mathrm{1}\right)^{\mathrm{2}{n}−\mathrm{2}} }{\left({m}−{j}\right)!\left({m}+{j}−\mathrm{1}\right)!} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}^{\mathrm{2}{m}} {x}}{{x}^{\mathrm{2}{n}−\mathrm{1}} }{dx}=\frac{\mathrm{4}^{{n}−{m}+\mathrm{1}} \left(\mathrm{2}{m}\right)!}{\mathrm{2}\left(\mathrm{2}{n}\right)!}\underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−{j}} {j}^{\mathrm{2}{n}} \mathrm{log}\:{j}}{\left({m}+{j}\right)!\left({m}−{j}\right)!} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}^{\mathrm{2}{m}−\mathrm{1}} {x}}{{x}^{\mathrm{2}{n}} }{dx}=\frac{\mathrm{4}^{\mathrm{1}−{m}} \left(\mathrm{2}{m}−\mathrm{1}\right)!}{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)!}\underset{{j}=\mathrm{1}} {\overset{{m}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−{j}} \left(\mathrm{2}{j}−\mathrm{1}\right)^{\mathrm{2}{n}−\mathrm{1}} \mathrm{log}\left(\mathrm{2}{j}−\mathrm{1}\right)}{\left({m}+{j}−\mathrm{1}\right)!\left({m}−{j}\right)!} \\ $$
Commented by MrGaster last updated on 19/Apr/25
