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Question Number 128314 by 676597498 last updated on 06/Jan/21

∫_0 ^( ∞) ((sin(px))/( (√x))) dx ∀p∈R

$$\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{sin}\left(\mathrm{px}\right)}{\:\sqrt{\mathrm{x}}}\:\mathrm{dx}\: \\ $$$$\forall\mathrm{p}\in\mathbb{R} \\ $$

Answered by Dwaipayan Shikari last updated on 06/Jan/21

∫_0 ^∞ ((sinpx)/( (√x)))dx=2∫_0 ^∞ sinpt^2 dt x=t^2 =(1/i)∫_0 ^∞ e^(ipt^2 ) −e^(−ipt^2 ) dt pt^2 =ij pt^2 =−ig =(1/(2p))∫_0 ^∞ (((ij)/p))^(−(1/2)) e^(−j) dj+(((−ig)/p))^(−(1/2)) e^(−g) dg =(1/(2(√p)))Γ((1/2))(e^((π/4)i) +e^(−(π/4)i) )=(√(π/(2p)))

$$\int_{\mathrm{0}} ^{\infty} \frac{{sinpx}}{\:\sqrt{{x}}}{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} {sinpt}^{\mathrm{2}} {dt}\:\:\:\:\:\:\:{x}={t}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{{i}}\int_{\mathrm{0}} ^{\infty} {e}^{{ipt}^{\mathrm{2}} } −{e}^{−{ipt}^{\mathrm{2}} } {dt}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{pt}^{\mathrm{2}} ={ij}\:\:\:\:{pt}^{\mathrm{2}} =−{ig} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{p}}\int_{\mathrm{0}} ^{\infty} \left(\frac{{ij}}{{p}}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−{j}} {dj}+\left(\frac{−{ig}}{{p}}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−{g}} {dg} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{p}}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left({e}^{\frac{\pi}{\mathrm{4}}{i}} +{e}^{−\frac{\pi}{\mathrm{4}}{i}} \right)=\sqrt{\frac{\pi}{\mathrm{2}{p}}} \\ $$

Answered by mnjuly1970 last updated on 06/Jan/21

solution: Ψ=∫_0 ^( ∞) ((sin(px))/x^(1/2) )dx Ψ=^(px=ξ) (1/p^(1/2) )∫_0 ^( ∞) ((sin(ξ))/ξ^(1/2) )dξ= =(1/p^(1/2) ) ∗(π/(2Γ((1/2))sin((π/4)))) =(π/( (√p) 2(√π) ((√2)/2))) = (π/( (√(2pπ)))) =((√(2pπ))/(2p))=(√(π/(2p)))

$${solution}:\:\Psi=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({px}\right)}{{x}^{\frac{\mathrm{1}}{\mathrm{2}}} }{dx} \\ $$$$\:\:\:\Psi\overset{{px}=\xi} {=}\frac{\mathrm{1}}{{p}^{\frac{\mathrm{1}}{\mathrm{2}}} }\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left(\xi\right)}{\xi^{\frac{\mathrm{1}}{\mathrm{2}}} }{d}\xi= \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{{p}^{\frac{\mathrm{1}}{\mathrm{2}}} }\:\ast\frac{\pi}{\mathrm{2}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right){sin}\left(\frac{\pi}{\mathrm{4}}\right)}\:=\frac{\pi}{\:\sqrt{{p}}\:\mathrm{2}\sqrt{\pi}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} \\ $$$$\:\:=\:\frac{\pi}{\:\sqrt{\mathrm{2}{p}\pi}}\:\:=\frac{\sqrt{\mathrm{2}{p}\pi}}{\mathrm{2}{p}}=\sqrt{\frac{\pi}{\mathrm{2}{p}}} \\ $$

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