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Question Number 146057 by mnjuly1970 last updated on 10/Jul/21

Answered by mathmax by abdo last updated on 10/Jul/21

J=Im(∫_0 ^∞ x^(−(1/2)) e^(−x+ix) dx) we have ∫_0 ^∞ x^(−(1/2)) e^((−1+i)x) dx =_((1−i)x=t) ∫_0 ^∞ ((t/(1−i)))^(−(1/2)) e^(−t) (dt/(1−i)) =(1/((1−i)^(1/2) ))∫_0 ^∞ t^(−(1/2)) e^(−t) dt =(1/(((√2)e^(−((iπ)/4)) )^(1/2) ))∫_0 ^∞ e^((1/2)−1) e^(−t) dt =Γ((1/2)).(1/((^4 (√2))))e^((iπ)/8) =((√π)/((^4 (√2)))){cos((π/8))+isin((π/8))} ⇒ J=((√π)/((^4 (√2))))sin((π/8)) ⇒λ=((√π)/((^4 (√2))))

$$\mathrm{J}=\mathrm{Im}\left(\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{e}^{−\mathrm{x}+\mathrm{ix}} \mathrm{dx}\right)\:\:\mathrm{we}\:\mathrm{have} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{e}^{\left(−\mathrm{1}+\mathrm{i}\right)\mathrm{x}} \:\mathrm{dx}\:=_{\left(\mathrm{1}−\mathrm{i}\right)\mathrm{x}=\mathrm{t}} \:\:\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{t}}{\mathrm{1}−\mathrm{i}}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{e}^{−\mathrm{t}} \frac{\mathrm{dt}}{\mathrm{1}−\mathrm{i}} \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{i}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{e}^{−\mathrm{t}} \:\mathrm{dt}\:=\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{t}} \:\mathrm{dt} \\ $$$$=\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right).\frac{\mathrm{1}}{\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{8}}} \:=\frac{\sqrt{\pi}}{\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)}\left\{\mathrm{cos}\left(\frac{\pi}{\mathrm{8}}\right)+\mathrm{isin}\left(\frac{\pi}{\mathrm{8}}\right)\right\}\:\Rightarrow \\ $$$$\mathrm{J}=\frac{\sqrt{\pi}}{\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)}\mathrm{sin}\left(\frac{\pi}{\mathrm{8}}\right)\:\Rightarrow\lambda=\frac{\sqrt{\pi}}{\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)} \\ $$

Commented by SANOGO last updated on 30/Aug/21

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