Question Number 220377 by SdC355 last updated on 12/May/25
$$\mathrm{Prove}\:\mathrm{equation} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:{f}\left({u}\right)\mathrm{g}\left({u}\right){e}^{−{u}\rho} \mathrm{d}{u}=\frac{\mathrm{1}}{\mathrm{2}\pi\boldsymbol{{i}}}\:\int_{−\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} ^{\:+\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} \:\:{F}\left({u}\right){G}\left({u}−\rho\right)\mathrm{d}{u} \\ $$$${F}\left({u}\right)=\int_{\mathrm{0}} ^{\:\infty} \:{f}\left({t}\right){e}^{−{ut}} \mathrm{d}{t} \\ $$$${G}\left({u}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{g}\left({t}\right){e}^{−{ut}} \mathrm{d}{t} \\ $$
Answered by MrGaster last updated on 12/May/25
![(1):∫_0 ^∞ f(u)g(u)c^(−uρ) =L{f(g)g(u)}(ρ) L{f(u)g(u)}(ρ)=(1/(2πi))∫_(γ−i∞) ^(γ+i∞) F(z)G(ρ−z)dz ⇒∫_0 ^∞ f(u)g(u)e^(−uρ) du=(1/(2πi))∫_(γ−i∞) ^(γ+i∞) F(z)G(ρ−z)dz z=u−ρ⇒dz=du ∫_(−∞i+γ) ^(+∞i+γ) F(u)G(u−ρ)du=∫_(γ−i∞) ^(γ+i∞) F(z+ρ)G(ρ−z)dz ⇒(1/(2πi))∫_(−∞i+γ) ^(+∞i+γ) F(u)G(u−ρ)du=L{f(u)g(u)}(ρ) ∫_0 ^( ∞) f(u)g(u)e^(−uρ) du=(1/(2πi)) ∫_(−∞i+𝛄) ^( +∞i+𝛄) F(u)G(u−ρ)du [Q.E.D] (2): { ((∫_0 ^( ∞) f(u)g(u)e^(−uρ) du=(1/(2πi)) ∫_(−∞i+𝛄) ^( +∞i+𝛄) F(u)G(u−ρ)du)),((F(u)=∫_0 ^( ∞) f(t)e^(−ut) dt)),((G(u)=∫_0 ^( ∞) g(t)e^(−ut) dt)) :} ∫_0 ^∞ (u)g(n)e^(−uρ) du=∫_0 ^∞ f(u)((1/(2πi))∫_(γ−i∞) ^(γ+i∞) G(s)e^(su) ds)e^(−up) du =(1/(2πi))∫_(γ−i∞) ^(γ+i∞) G(s)∫_0 ^∞ f(u)e^(−(ρ−s)) du ds =(1/(2πi))∫_(γ−i∞) ^(γ+i∞) G(s)∫_0 ^∞ f(u)e^(−(ρ−s)) du ds =(1/(2πi))∫_(γ−i∞) ^(γ+i∞) G(s)F(ρ−s)ds =(1/(2πi))∫_(γ−i∞) ^(γ+i∞) F(u)G(ρ−u)du (s=ρ−u) ∫_0 ^( ∞) f(u)g(u)e^(−uρ) du=(1/(2πi)) ∫_(−∞i+𝛄) ^( +∞i+𝛄) F(u)G(u−ρ)du [Q.E.D] (3)∫_0 ^( ∞) f(u)g(u)e^(−uρ) du=∫_0 ^∞ ((1/(2πi))∫_(γ−i∞) ^(γ+i∞) F(s)e^(su) ds)((1/(2πi))∫_(γ′−i∞) ^(γ′+i∞) G(s′)e^(s′u) ds′)e^(−uρ) du =(1/((2πi)^2 ))∫_0 ^∞ ∫_(γ−i∞) ^(γ+i∞) ∫_(γ′−i∞) ^(γ′+i∞) F(s)G(s′)e^(u(s+s′−ρ)) ds ds′du =(1/((2πi)^2 ))∫_(γ−i∞) ^(γ+i∞) ∫_(γ′−i∞) ^(γ′+i∞) F(s)G(s′)(∫_0 ^∞ e^(u(s+s′−ρ)) )ds ds′ =(1/((2πi)^2 ))∫_(γ−i∞) ^(γ+i∞) ∫_(γ′−i∞) ^(γ′+i∞) ((F(s)G(s′))/(ρ−s−s′))ds ds′ =(1/(2πi))∫_(γ−i∞) ^(γ+i∞) F(u)G(ρ−u)du (by Parseval′s theorem) =(1/(2πi))∫_(−∞i+γ) ^(+∞i+γ) F(u)G(u−ρ)du determinant (((Proven)))](https://www.tinkutara.com/question/Q220455.png)
$$\left(\mathrm{1}\right):\int_{\mathrm{0}} ^{\infty} {f}\left({u}\right){g}\left({u}\right){c}^{−{u}\rho} =\mathcal{L}\left\{{f}\left({g}\right){g}\left({u}\right)\right\}\left(\rho\right) \\ $$$$\mathcal{L}\left\{{f}\left({u}\right){g}\left({u}\right)\right\}\left(\rho\right)=\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\int_{\gamma−{i}\infty} ^{\gamma+{i}\infty} {F}\left({z}\right){G}\left(\rho−{z}\right){dz} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} {f}\left({u}\right){g}\left({u}\right){e}^{−{u}\rho} {du}=\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\int_{\gamma−{i}\infty} ^{\gamma+{i}\infty} {F}\left({z}\right){G}\left(\rho−{z}\right){dz} \\ $$$${z}={u}−\rho\Rightarrow{dz}={du} \\ $$$$\int_{−\infty{i}+\gamma} ^{+\infty{i}+\gamma} {F}\left({u}\right){G}\left({u}−\rho\right){du}=\int_{\gamma−{i}\infty} ^{\gamma+{i}\infty} {F}\left({z}+\rho\right){G}\left(\rho−{z}\right){dz} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\int_{−\infty{i}+\gamma} ^{+\infty{i}+\gamma} {F}\left({u}\right){G}\left({u}−\rho\right){du}=\mathcal{L}\left\{{f}\left({u}\right){g}\left({u}\right)\right\}\left(\rho\right) \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:{f}\left({u}\right)\mathrm{g}\left({u}\right){e}^{−{u}\rho} \mathrm{d}{u}=\frac{\mathrm{1}}{\mathrm{2}\pi\boldsymbol{{i}}}\:\int_{−\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} ^{\:+\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} \:\:{F}\left({u}\right){G}\left({u}−\rho\right)\mathrm{d}{u} \\ $$$$\left[\mathrm{Q}.\mathrm{E}.\mathrm{D}\right] \\ $$$$\left(\mathrm{2}\right):\begin{cases}{\int_{\mathrm{0}} ^{\:\infty} \:{f}\left({u}\right)\mathrm{g}\left({u}\right){e}^{−{u}\rho} \mathrm{d}{u}=\frac{\mathrm{1}}{\mathrm{2}\pi\boldsymbol{{i}}}\:\int_{−\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} ^{\:+\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} \:\:{F}\left({u}\right){G}\left({u}−\rho\right)\mathrm{d}{u}}\\{{F}\left({u}\right)=\int_{\mathrm{0}} ^{\:\infty} \:{f}\left({t}\right){e}^{−{ut}} \mathrm{d}{t}}\\{{G}\left({u}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{g}\left({t}\right){e}^{−{ut}} \mathrm{d}{t}}\end{cases} \\ $$$$\int_{\mathrm{0}} ^{\infty} \left({u}\right){g}\left({n}\right){e}^{−{u}\rho} {du}=\int_{\mathrm{0}} ^{\infty} {f}\left({u}\right)\left(\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\int_{\gamma−{i}\infty} ^{\gamma+{i}\infty} {G}\left({s}\right){e}^{{su}} {ds}\right){e}^{−{up}} {du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\int_{\gamma−{i}\infty} ^{\gamma+{i}\infty} {G}\left({s}\right)\int_{\mathrm{0}} ^{\infty} {f}\left({u}\right){e}^{−\left(\rho−{s}\right)} {du}\:{ds} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\int_{\gamma−{i}\infty} ^{\gamma+{i}\infty} {G}\left({s}\right)\int_{\mathrm{0}} ^{\infty} {f}\left({u}\right){e}^{−\left(\rho−{s}\right)} {du}\:{ds} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\int_{\gamma−{i}\infty} ^{\gamma+{i}\infty} {G}\left({s}\right){F}\left(\rho−{s}\right){ds} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\int_{\gamma−{i}\infty} ^{\gamma+{i}\infty} {F}\left({u}\right){G}\left(\rho−{u}\right){du}\:\left({s}=\rho−{u}\right) \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:{f}\left({u}\right)\mathrm{g}\left({u}\right){e}^{−{u}\rho} \mathrm{d}{u}=\frac{\mathrm{1}}{\mathrm{2}\pi\boldsymbol{{i}}}\:\int_{−\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} ^{\:+\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} \:\:{F}\left({u}\right){G}\left({u}−\rho\right)\mathrm{d}{u} \\ $$$$\left[\mathrm{Q}.\mathrm{E}.\mathrm{D}\right] \\ $$$$\left(\mathrm{3}\right)\int_{\mathrm{0}} ^{\:\infty} \:{f}\left({u}\right)\mathrm{g}\left({u}\right){e}^{−{u}\rho} \mathrm{d}{u}=\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\int_{\gamma−{i}\infty} ^{\gamma+{i}\infty} {F}\left({s}\right){e}^{{su}} {ds}\right)\left(\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\int_{\gamma'−{i}\infty} ^{\gamma'+{i}\infty} {G}\left({s}'\right){e}^{{s}'{u}} {ds}'\right){e}^{−{u}\rho} {du} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{2}\pi{i}\right)^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \int_{\gamma−{i}\infty} ^{\gamma+{i}\infty} \int_{\gamma'−{i}\infty} ^{\gamma'+{i}\infty} {F}\left({s}\right){G}\left({s}'\right){e}^{{u}\left({s}+{s}'−\rho\right)} {ds}\:{ds}'{du} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{2}\pi{i}\right)^{\mathrm{2}} }\int_{\gamma−{i}\infty} ^{\gamma+{i}\infty} \int_{\gamma'−{i}\infty} ^{\gamma'+{i}\infty} {F}\left({s}\right){G}\left({s}'\right)\left(\int_{\mathrm{0}} ^{\infty} {e}^{{u}\left({s}+{s}'−\rho\right)} \right){ds}\:{d}\mathrm{s}' \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{2}\pi{i}\right)^{\mathrm{2}} }\int_{\gamma−{i}\infty} ^{\gamma+{i}\infty} \int_{\gamma'−{i}\infty} ^{\gamma'+{i}\infty} \frac{{F}\left({s}\right){G}\left({s}'\right)}{\rho−{s}−{s}'}{ds}\:{ds}' \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\int_{\gamma−{i}\infty} ^{\gamma+{i}\infty} {F}\left({u}\right){G}\left(\rho−{u}\right){du}\:\left(\mathrm{by}\:\mathrm{Parseval}'\mathrm{s}\:\mathrm{theorem}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\int_{−\infty{i}+\gamma} ^{+\infty{i}+\gamma} {F}\left({u}\right){G}\left({u}−\rho\right){du}\:\begin{array}{|c|}{\mathrm{Proven}}\\\hline\end{array} \\ $$