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Question Number 135309 by mnjuly1970 last updated on 12/Mar/21
.... Nice Calculus .... prove that ::: 𝛗=∫_0 ^( ∞) ((sin(ksinα)x)/( (√x)))dx=(√(π/k)) sin((α/2))...
$$\:\:\:\:\:\:\:\:\:\:….\:\:\mathscr{N}{ice}\:\:\:\:\mathscr{C}{alculus}\:…. \\ $$$$\:\:\:\:\:\:\:\:{prove}\:\:{that}\:::: \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({ksin}\alpha\right){x}}{\:\sqrt{{x}}}{dx}=\sqrt{\frac{\pi}{{k}}}\:{sin}\left(\frac{\alpha}{\mathrm{2}}\right)… \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\: \\ $$
Answered by mathmax by abdo last updated on 12/Mar/21
Φ =∫_0 ^∞ ((sin(ksinα)x)/( (√x))) we put ksinα=λ ⇒Φ=∫_0 ^∞ ((sin(λx))/( (√x)))dx =−Im(∫_0 ^∞ (e^(−iλx) /( (√x)))dx) we have ∫_0 ^∞ (e^(−iλx) /( (√x)))dx =_((√x)=t) ∫_0 ^∞ (e^(−iλt^2 ) /t)(2t)dt =2∫_0 ^∞ e^(−iλt^2 ) dt =2 ∫_0 ^∞ e^(−((√(λi))t)^2 ) dt =_((√(λi))t=z) 2∫_0 ^∞ e^(−z^2 ) (dz/( (√(λi)))) =(1/( (√i)(√λ)))×2.((√π)/2) =(√(π/λ))e^(−((iπ)/4)) =(√(π/λ)){(1/( (√2)))−(i/( (√2)))} ⇒ Φ=(1/( (√2)))(√(π/λ))=(1/( (√2)))(√(π/(ksinα)))=(√(π/(2ksinα)))
$$\Phi\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}\left(\mathrm{ksin}\alpha\right)\mathrm{x}}{\:\sqrt{\mathrm{x}}}\:\:\mathrm{we}\:\mathrm{put}\:\mathrm{ksin}\alpha=\lambda\:\Rightarrow\Phi=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}\left(\lambda\mathrm{x}\right)}{\:\sqrt{\mathrm{x}}}\mathrm{dx} \\ $$$$=−\mathrm{Im}\left(\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{e}^{−\mathrm{i}\lambda\mathrm{x}} }{\:\sqrt{\mathrm{x}}}\mathrm{dx}\right)\:\:\mathrm{we}\:\mathrm{have}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{e}^{−\mathrm{i}\lambda\mathrm{x}} }{\:\sqrt{\mathrm{x}}}\mathrm{dx}\:=_{\sqrt{\mathrm{x}}=\mathrm{t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{e}^{−\mathrm{i}\lambda\mathrm{t}^{\mathrm{2}} } }{\mathrm{t}}\left(\mathrm{2t}\right)\mathrm{dt} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{i}\lambda\mathrm{t}^{\mathrm{2}} } \mathrm{dt}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{e}^{−\left(\sqrt{\lambda\mathrm{i}}\mathrm{t}\right)^{\mathrm{2}} } \mathrm{dt}\:=_{\sqrt{\lambda\mathrm{i}}\mathrm{t}=\mathrm{z}} \:\:\:\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{z}^{\mathrm{2}} } \frac{\mathrm{dz}}{\:\sqrt{\lambda\mathrm{i}}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{i}}\sqrt{\lambda}}×\mathrm{2}.\frac{\sqrt{\pi}}{\mathrm{2}}\:\:=\sqrt{\frac{\pi}{\lambda}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} =\sqrt{\frac{\pi}{\lambda}}\left\{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}\right\}\:\Rightarrow \\ $$$$\Phi=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\sqrt{\frac{\pi}{\lambda}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\sqrt{\frac{\pi}{\mathrm{ksin}\alpha}}=\sqrt{\frac{\pi}{\mathrm{2ksin}\alpha}} \\ $$$$ \\ $$

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