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Question Number 97563 by 675480065 last updated on 08/Jun/20

∫_0 ^(+∞) e^(−x) cos(x^2 )dx.  Discuss the convergence of this   generalised intergral.  Please help

$$\int_{\mathrm{0}} ^{+\infty} \mathrm{e}^{−\mathrm{x}} \mathrm{cos}\left(\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx}. \\ $$$$\mathrm{Discuss}\:\mathrm{the}\:\mathrm{convergence}\:\mathrm{of}\:\mathrm{this}\: \\ $$$$\mathrm{generalised}\:\mathrm{intergral}. \\ $$$$\mathrm{Please}\:\mathrm{help} \\ $$

Answered by mathmax by abdo last updated on 08/Jun/20

I =∫_0 ^∞  e^(−x)  cos(x^2 )dx   for all a>0  x→e^(−x)  cos(x^2 )is continue on [0,a] so  integrable on[0,a]  at [a,+∞[  lim_(x→+∞) e^(−x)  cos(x^2 )dx =0 ⇒  ∫_a ^(+∞)  e^(−x)  cos(x^2 )dx cv ⇒∫_0 ^∞  e^(−x)  cos(x^2 )dx converge.  the vslue  I =Re( ∫_0 ^∞ e^(−x+ix^2 ) dx) and   ∫_0 ^∞  e^(−x+ix^2 ) dx =∫_0 ^∞  e^(((√i)x)^2 −2(√i)x×(1/(2(√i)))  +((1/(2(√i))))^2 −((1/(2(√i))))^2 ) dx  =∫_0 ^∞    e^(((√i)x−(1/(2(√i))))^2 −(1/(4i)))  dx =e^(i/4)  ∫_0 ^∞  e^(((√i)x−(1/(2(√i))))^2 ) dx =_((√i)x−(1/(2(√i)))=t)   e^(i/4)  ∫_(−(1/(2(√i)))) ^(+∞)  e^(−t^2 ) (dt/(√i))  =e^(−((iπ)/4))  e^(i/4)  ∫_(−(1/2)e^(−((iπ)/4)) ) ^(+∞)  e^(−t^2 ) dt  =e^((−(π/4)+(1/4))i)  {((√π)/2)−∫_0 ^(−(1/2)e^(−t^2 ) ) dt}...be continued...

$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{x}} \:\mathrm{cos}\left(\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx}\:\:\:\mathrm{for}\:\mathrm{all}\:\mathrm{a}>\mathrm{0}\:\:\mathrm{x}\rightarrow\mathrm{e}^{−\mathrm{x}} \:\mathrm{cos}\left(\mathrm{x}^{\mathrm{2}} \right)\mathrm{is}\:\mathrm{continue}\:\mathrm{on}\:\left[\mathrm{0},\mathrm{a}\right]\:\mathrm{so} \\ $$$$\mathrm{integrable}\:\mathrm{on}\left[\mathrm{0},\mathrm{a}\right]\:\:\mathrm{at}\:\left[\mathrm{a},+\infty\left[\:\:\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \mathrm{e}^{−\mathrm{x}} \:\mathrm{cos}\left(\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx}\:=\mathrm{0}\:\Rightarrow\right.\right. \\ $$$$\int_{\mathrm{a}} ^{+\infty} \:\mathrm{e}^{−\mathrm{x}} \:\mathrm{cos}\left(\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx}\:\mathrm{cv}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{x}} \:\mathrm{cos}\left(\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx}\:\mathrm{converge}. \\ $$$$\mathrm{the}\:\mathrm{vslue}\:\:\mathrm{I}\:=\mathrm{Re}\left(\:\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{x}+\mathrm{ix}^{\mathrm{2}} } \mathrm{dx}\right)\:\mathrm{and}\: \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{x}+\mathrm{ix}^{\mathrm{2}} } \mathrm{dx}\:=\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{\left(\sqrt{\mathrm{i}}\mathrm{x}\right)^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{i}}\mathrm{x}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{i}}}\:\:+\left(\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{i}}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{i}}}\right)^{\mathrm{2}} } \mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\mathrm{e}^{\left(\sqrt{\mathrm{i}}\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{i}}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4i}}} \:\mathrm{dx}\:=\mathrm{e}^{\frac{\mathrm{i}}{\mathrm{4}}} \:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{\left(\sqrt{\mathrm{i}}\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{i}}}\right)^{\mathrm{2}} } \mathrm{dx}\:=_{\sqrt{\mathrm{i}}\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{i}}}=\mathrm{t}} \:\:\mathrm{e}^{\frac{\mathrm{i}}{\mathrm{4}}} \:\int_{−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{i}}}} ^{+\infty} \:\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \frac{\mathrm{dt}}{\sqrt{\mathrm{i}}} \\ $$$$=\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \:\mathrm{e}^{\frac{\mathrm{i}}{\mathrm{4}}} \:\int_{−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} } ^{+\infty} \:\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \mathrm{dt}\:\:=\mathrm{e}^{\left(−\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}\right)\mathrm{i}} \:\left\{\frac{\sqrt{\pi}}{\mathrm{2}}−\int_{\mathrm{0}} ^{−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } } \mathrm{dt}\right\}...\mathrm{be}\:\mathrm{continued}... \\ $$

Commented by 675480065 last updated on 08/Jun/20

thanks teacher

$$\boldsymbol{{thanks}}\:\boldsymbol{{teacher}} \\ $$

Commented by abdomathmax last updated on 08/Jun/20

you are welcome.

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

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