Question Number 222217 by MrGaster last updated on 20/Jun/25
![∫_0 ^∞ ∫_0 ^∞ ((ln x ln y)/( (√(xy))))cos(x+y)=π^2 (γ+2 ln 2) Sol:∫_0 ^∞ ∫_0 ^∞ ((ln x ln y)/( (√(xy))))cos(x+y)dxdy=Re((∫_0 ^∞ x^(−(1/2)) e^(ix) ln xdx)(∫_0 ^∞ y^(−(1/2)) e^(iy) ln ydy)) ∫_0 ^∞ x^a e^(ix) dx=e^(iπ(a+1)) Γ(a+1),−1<Re a<0 ∫_0 ^∞ x^a e^(ix) dx=∫_0 ^∞ x^a e^(ix) ln xdx=(∂/∂u)[e^(iπ(a+1)/2) Γ(a+1)] =e^(iπ(a+1)) Γ(a+1)(((iπ)/2)+ψ(a+1)) a=−(1/2): ∫_0 ^∞ x^(−(1/2)) e^(ix) ln xdx=e^(iπ/1) (√π)(((iπ)/2)+ψ((1/2))) c=e^(iπ/4) (√π)(((iπ)/2)−γ−2 ln 2) e^(iπ/4) =((√2)/2)(1+i) c=(√π)∙((√2)/2)(1+i)(−γ−2 ln 2+i(π/2))=((√(2π))/2)[(γ−ln 2−(π/2))+i(−γ−2 ln 2+(π/2))] p=−γ−2 ln 2−(π/2),q=−γ−2 ln 2+(π/2) c^2 =(((√(2π))/2))^2 (p+ip)^2 =((2π)/4)(p^2 −q^2 +2ipq)=(π/2)(p^2 −q^2 +2ipq) p+q+2=(−γ−2 ln 2) p−q=π p^2 −q^2 =(p−q)(p+q)=(−π)∙2(−γ−2 ln 2)=2π(γ+2 ln 2) [2π(γ+2 ln 2)+2ipq]=π^2 (γ+2 ln 2)+iπpq Re(c^2 )=π^2 (γ+2 ln 2)](https://www.tinkutara.com/question/Q222217.png)
$$\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\:{x}\:\mathrm{ln}\:{y}}{\:\sqrt{{xy}}}\mathrm{cos}\left({x}+{y}\right)=\pi^{\mathrm{2}} \left(\gamma+\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right) \\ $$$$\mathrm{Sol}:\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\:{x}\:\mathrm{ln}\:{y}}{\:\sqrt{{xy}}}\mathrm{cos}\left({x}+{y}\right){dxdy}=\mathrm{Re}\left(\left(\int_{\mathrm{0}} ^{\infty} {x}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{{ix}} \mathrm{ln}\:{xdx}\right)\left(\int_{\mathrm{0}} ^{\infty} {y}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{{iy}} \mathrm{ln}\:{ydy}\right)\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} {x}^{{a}} {e}^{{ix}} {dx}={e}^{{i}\pi\left({a}+\mathrm{1}\right)} \Gamma\left({a}+\mathrm{1}\right),−\mathrm{1}<\mathrm{Re}\:{a}<\mathrm{0} \\ $$$$\int_{\mathrm{0}} ^{\infty} {x}^{{a}} {e}^{{ix}} {dx}=\int_{\mathrm{0}} ^{\infty} {x}^{{a}} {e}^{{ix}} \mathrm{ln}\:{xdx}=\frac{\partial}{\partial{u}}\left[{e}^{{i}\pi\left({a}+\mathrm{1}\right)/\mathrm{2}} \Gamma\left({a}+\mathrm{1}\right)\right] \\ $$$$={e}^{{i}\pi\left({a}+\mathrm{1}\right)} \Gamma\left({a}+\mathrm{1}\right)\left(\frac{{i}\pi}{\mathrm{2}}+\psi\left({a}+\mathrm{1}\right)\right) \\ $$$${a}=−\frac{\mathrm{1}}{\mathrm{2}}: \\ $$$$\int_{\mathrm{0}} ^{\infty} {x}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{{ix}} \mathrm{ln}\:{xdx}={e}^{{i}\pi/\mathrm{1}} \sqrt{\pi}\left(\frac{{i}\pi}{\mathrm{2}}+\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$${c}={e}^{{i}\pi/\mathrm{4}} \sqrt{\pi}\left(\frac{{i}\pi}{\mathrm{2}}−\gamma−\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right) \\ $$$${e}^{{i}\pi/\mathrm{4}} =\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}+{i}\right) \\ $$$${c}=\sqrt{\pi}\centerdot\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}+{i}\right)\left(−\gamma−\mathrm{2}\:\mathrm{ln}\:\mathrm{2}+{i}\frac{\pi}{\mathrm{2}}\right)=\frac{\sqrt{\mathrm{2}\pi}}{\mathrm{2}}\left[\left(\gamma−\mathrm{ln}\:\mathrm{2}−\frac{\pi}{\mathrm{2}}\right)+{i}\left(−\gamma−\mathrm{2}\:\mathrm{ln}\:\mathrm{2}+\frac{\pi}{\mathrm{2}}\right)\right] \\ $$$${p}=−\gamma−\mathrm{2}\:\mathrm{ln}\:\mathrm{2}−\frac{\pi}{\mathrm{2}},{q}=−\gamma−\mathrm{2}\:\mathrm{ln}\:\mathrm{2}+\frac{\pi}{\mathrm{2}} \\ $$$${c}^{\mathrm{2}} =\left(\frac{\sqrt{\mathrm{2}\pi}}{\mathrm{2}}\right)^{\mathrm{2}} \left({p}+{ip}\right)^{\mathrm{2}} =\frac{\mathrm{2}\pi}{\mathrm{4}}\left({p}^{\mathrm{2}} −{q}^{\mathrm{2}} +\mathrm{2}{ipq}\right)=\frac{\pi}{\mathrm{2}}\left({p}^{\mathrm{2}} −{q}^{\mathrm{2}} +\mathrm{2}{ipq}\right) \\ $$$${p}+{q}+\mathrm{2}=\left(−\gamma−\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right) \\ $$$${p}−{q}=\pi \\ $$$${p}^{\mathrm{2}} −{q}^{\mathrm{2}} =\left({p}−{q}\right)\left({p}+{q}\right)=\left(−\pi\right)\centerdot\mathrm{2}\left(−\gamma−\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right)=\mathrm{2}\pi\left(\gamma+\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right) \\ $$$$\left[\mathrm{2}\pi\left(\gamma+\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right)+\mathrm{2}{ipq}\right]=\pi^{\mathrm{2}} \left(\gamma+\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right)+{i}\pi{pq} \\ $$$$\mathrm{Re}\left({c}^{\mathrm{2}} \right)=\pi^{\mathrm{2}} \left(\gamma+\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right) \\ $$