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Question Number 221350 by MrGaster last updated on 31/May/25
∫_0 ^∞ ((cos πx)/(Γ(2+x)Γ(2−x)))dx
$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{cos}\:\pi{x}}{\Gamma\left(\mathrm{2}+{x}\right)\Gamma\left(\mathrm{2}−{x}\right)}{dx} \\ $$$$ \\ $$$$ \\ $$
Commented by Ghisom last updated on 31/May/25
Γ (2+x) Γ (2−x) =((πx(1−x)(1+x))/(sin πx)) ∫_0 ^∞ ((cos πx)/(Γ (2+x) Γ (2−x)))dx= =(1/(2π))∫_0 ^∞ ((sin 2πx)/(x(1−x^2 )))dx= =2π∫_0 ^∞ ((sin t)/(t(2π−t)(2π+t)))dt= =(1/(2π))∫_0 ^∞ ((sin t)/t)dt−(1/(4π))∫_0 ^∞ ((sin t)/(t−2π))dt−(1/(4π))∫_0 ^∞ ((sin t)/(t+2π))dt ⇒ the answer should be 0
$$\Gamma\:\left(\mathrm{2}+{x}\right)\:\Gamma\:\left(\mathrm{2}−{x}\right)\:=\frac{\pi{x}\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}{\mathrm{sin}\:\pi{x}} \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{cos}\:\pi{x}}{\Gamma\:\left(\mathrm{2}+{x}\right)\:\Gamma\:\left(\mathrm{2}−{x}\right)}{dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{sin}\:\mathrm{2}\pi{x}}{{x}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{dx}= \\ $$$$=\mathrm{2}\pi\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{sin}\:{t}}{{t}\left(\mathrm{2}\pi−{t}\right)\left(\mathrm{2}\pi+{t}\right)}{dt}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{sin}\:{t}}{{t}}{dt}−\frac{\mathrm{1}}{\mathrm{4}\pi}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{sin}\:{t}}{{t}−\mathrm{2}\pi}{dt}−\frac{\mathrm{1}}{\mathrm{4}\pi}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{sin}\:{t}}{{t}+\mathrm{2}\pi}{dt} \\ $$$$\Rightarrow \\ $$$$\mathrm{the}\:\mathrm{answer}\:\mathrm{should}\:\mathrm{be}\:\mathrm{0} \\ $$
Answered by SdC355 last updated on 31/May/25
∫_0 ^( ∞) ((cos(πt))/(Γ(2+t)Γ(2−t))) dt=−(1/π) ∫_0 ^( ∞) ((sin(πt)cos(πt))/(t(t^2 −1))) dt −(1/(2π)) ∫_0 ^( ∞) ((sin(2πt))/(t(t^2 −1))) dt sin(2πt)=((e^(2iπt) −e^(−2iπt) )/(2i)) I=J{∫_0 ^( ∞) (e^(2iπt) /(t(t^2 −1))) dt} C=C_R +Arc_R ∣∫_(Arc_R ) ^ f(z)dz∣→0 as R→∞ Total Residue S=Σ_h Res_(z=z_h ) {(e^(2iπt) /(t(t^2 −1)))}=0 ∮_C f(z)dz=2πi∙0=0 ∫_(−∞) ^(+∞) =∫_0 ^( ∞) =0 ∴∫_0 ^( ∞) ((sin(2πt))/(t(t^2 −1))) dt=0
$$\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{cos}\left(\pi{t}\right)}{\Gamma\left(\mathrm{2}+{t}\right)\Gamma\left(\mathrm{2}−{t}\right)}\:\mathrm{d}{t}=−\frac{\mathrm{1}}{\pi}\:\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{sin}\left(\pi{t}\right)\mathrm{cos}\left(\pi{t}\right)}{{t}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}\:\mathrm{d}{t} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}\pi}\:\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{sin}\left(\mathrm{2}\pi{t}\right)}{{t}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}\:\mathrm{d}{t} \\ $$$$\mathrm{sin}\left(\mathrm{2}\pi{t}\right)=\frac{{e}^{\mathrm{2}\boldsymbol{{i}}\pi{t}} −{e}^{−\mathrm{2}\boldsymbol{{i}}\pi{t}} }{\mathrm{2}\boldsymbol{{i}}} \\ $$$${I}=\boldsymbol{\mathfrak{J}}\left\{\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{{e}^{\mathrm{2}\boldsymbol{{i}}\pi{t}} }{{t}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}\:\mathrm{d}{t}\right\} \\ $$$${C}={C}_{{R}} +\mathrm{Arc}_{{R}} \\ $$$$\mid\int_{\mathrm{Arc}_{{R}} } ^{\:} \:{f}\left({z}\right)\mathrm{d}{z}\mid\rightarrow\mathrm{0}\:\:\mathrm{as}\:{R}\rightarrow\infty \\ $$$$\mathrm{Total}\:\mathrm{Residue}\:{S}=\underset{{h}} {\sum}\:\mathrm{Res}_{{z}={z}_{{h}} } \left\{\frac{{e}^{\mathrm{2}\boldsymbol{{i}}\pi{t}} }{{t}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}\right\}=\mathrm{0} \\ $$$$\oint_{{C}} \:{f}\left({z}\right)\mathrm{d}{z}=\mathrm{2}\pi\boldsymbol{{i}}\centerdot\mathrm{0}=\mathrm{0} \\ $$$$\int_{−\infty} ^{+\infty} \:=\int_{\mathrm{0}} ^{\:\infty} =\mathrm{0} \\ $$$$\therefore\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{sin}\left(\mathrm{2}\pi{t}\right)}{{t}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}\:\mathrm{d}{t}=\mathrm{0} \\ $$

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