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Question Number 82867 by mind is power last updated on 24/Feb/20

hello prove that ∫_0 ^(+∞) sin(x^4 )dx=sin((π/8))∫_0 ^(+∞) e^(−x^4 ) dx? verry nice day Good Bless You

$${hello}\:{prove}\:{that}\:\int_{\mathrm{0}} ^{+\infty} {sin}\left({x}^{\mathrm{4}} \right){dx}={sin}\left(\frac{\pi}{\mathrm{8}}\right)\int_{\mathrm{0}} ^{+\infty} {e}^{−{x}^{\mathrm{4}} } {dx}? \\ $$$${verry}\:{nice}\:{day}\:{Good}\:{Bless}\:{You} \\ $$

Commented by abdomathmax last updated on 25/Feb/20

∫_0 ^∞ sin(x^4 )dx =Im(∫_0 ^∞ e^(ix^4 ) dx) but chsngement ix^4 =−t^4 give x^4 =it^4 ⇒x =i^(1/4) t ⇒ ∫_0 ^∞ e^(ix^4 ) dx =∫_0 ^∞ e^(−t^4 ) i^(1/4) dt =(e^((iπ)/2) )^(1/4) ∫_0 ^∞ e^(−t^4 ) dt =e^((iπ)/8) ∫_0 ^∞ e^(−t^4 ) dt =(cos((π/8))+isin((π/8)))∫_0 ^∞ e^(−t^4 ) dt ⇒ ∫_0 ^∞ sin(x^4 )dx =sin((π/8))∫_0 ^∞ e^(−t^4 ) dt also chang.t^4 =u give t =u^(1/4) ⇒ ∫_0 ^∞ e^(−t^4 ) dt =∫_0 ^∞ e^(−u) (1/4)u^((1/4)−1) du =(1/4)Γ((1/4))⇒∫_0 ^∞ sin(x^4 )dx =(1/4)sin((π/8))Γ((1/4)) =((√(2−(√2)))/8)×Γ((1/4))

$$\int_{\mathrm{0}} ^{\infty} \:{sin}\left({x}^{\mathrm{4}} \right){dx}\:={Im}\left(\int_{\mathrm{0}} ^{\infty} {e}^{{ix}^{\mathrm{4}} } {dx}\right)\:{but}\:\:{chsngement} \\ $$$${ix}^{\mathrm{4}} =−{t}^{\mathrm{4}} \:{give}\:{x}^{\mathrm{4}} ={it}^{\mathrm{4}} \:\Rightarrow{x}\:={i}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{t}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{{ix}^{\mathrm{4}} } {dx}\:=\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{t}^{\mathrm{4}} } {i}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{dt}\:=\left({e}^{\frac{{i}\pi}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}^{\mathrm{4}} } {dt} \\ $$$$={e}^{\frac{{i}\pi}{\mathrm{8}}} \:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}^{\mathrm{4}} } {dt}\:\:=\left({cos}\left(\frac{\pi}{\mathrm{8}}\right)+{isin}\left(\frac{\pi}{\mathrm{8}}\right)\right)\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}^{\mathrm{4}} } {dt}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{sin}\left({x}^{\mathrm{4}} \right){dx}\:={sin}\left(\frac{\pi}{\mathrm{8}}\right)\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}^{\mathrm{4}} } {dt}\: \\ $$$${also}\:{chang}.{t}^{\mathrm{4}} ={u}\:{give}\:{t}\:={u}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}^{\mathrm{4}} } {dt}\:=\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{u}} \frac{\mathrm{1}}{\mathrm{4}}{u}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} {du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\Rightarrow\int_{\mathrm{0}} ^{\infty} \:{sin}\left({x}^{\mathrm{4}} \right){dx}\:=\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\frac{\pi}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$=\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{8}}×\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$

Commented by mind is power last updated on 25/Feb/20

nice solution Sir thank you

$${nice}\:{solution}\:{Sir}\:{thank}\:{you} \\ $$

Commented by msup trace by abdo last updated on 25/Feb/20

you are welcome

$${you}\:{are}\:{welcome} \\ $$

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