Verify that $left|int_{gamma} exp(iz^2)dzright| leq frac{pibig(1-exp(-r^2)big)}{4r}$ where...












2















Verify that $$left|int_{gamma} exp(iz^2)dzright| leq frac{pibig(1-exp(-r^2)big)}{4r}$$ where $gamma(t)=re^{it}$, for $0leq t leq pi/4$ and $r > 0$.




I'm stuck. here is my attempt:



$|int_{gamma} e^{icdot z^2}dz|leq int_{gamma} |e^{icdot z^2}|dz=int_{gamma}|(e^{z^2})^i|dz$. As $t > 0$ on $0leq t leq pi/4$.
$Rightarrow|e^{r^2 e^{2it}}|=|e^{r^2}e^{e^{2it}}|>0$
$Rightarrowint_{0}^{pi/4} e^{r^2}e^{2it}rie^{it}dt=e^{r^2}riint_{0}^{pi/4} exp(e^{2it}+it)dt$

Let $alpha=e^{r^2}ri$
$Rightarrow alphaint_{0}^{pi/4}e^{cos2t}e^{icdot sin2t}(cos(t)+icdot sin(t)dt)$



Am I on track?






integration complex-analysis contour-integration integral-inequality holomorphic-functions







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  • 1




    What is $sen(t)$? Did you mean to write $sin(t)$?
    – LoveTooNap29
    yesterday










  • yeah, I corrected that. tks
    – Lincon Ribeiro
    yesterday






  • 1




    Hints: $|e^{i(x+iy)^2}|=e^{-2xy}$ and $sinphigeqslant 2phi/pi$ for $0leqslantphileqslantpi/2$.
    – metamorphy
    yesterday


















2















Verify that $$left|int_{gamma} exp(iz^2)dzright| leq frac{pibig(1-exp(-r^2)big)}{4r}$$ where $gamma(t)=re^{it}$, for $0leq t leq pi/4$ and $r > 0$.




I'm stuck. here is my attempt:



$|int_{gamma} e^{icdot z^2}dz|leq int_{gamma} |e^{icdot z^2}|dz=int_{gamma}|(e^{z^2})^i|dz$. As $t > 0$ on $0leq t leq pi/4$.
$Rightarrow|e^{r^2 e^{2it}}|=|e^{r^2}e^{e^{2it}}|>0$
$Rightarrowint_{0}^{pi/4} e^{r^2}e^{2it}rie^{it}dt=e^{r^2}riint_{0}^{pi/4} exp(e^{2it}+it)dt$

Let $alpha=e^{r^2}ri$
$Rightarrow alphaint_{0}^{pi/4}e^{cos2t}e^{icdot sin2t}(cos(t)+icdot sin(t)dt)$



Am I on track?






integration complex-analysis contour-integration integral-inequality holomorphic-functions







share|cite|improve this question









New contributor




Lincon Ribeiro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 1




    What is $sen(t)$? Did you mean to write $sin(t)$?
    – LoveTooNap29
    yesterday










  • yeah, I corrected that. tks
    – Lincon Ribeiro
    yesterday






  • 1




    Hints: $|e^{i(x+iy)^2}|=e^{-2xy}$ and $sinphigeqslant 2phi/pi$ for $0leqslantphileqslantpi/2$.
    – metamorphy
    yesterday
















2












2








2


0






Verify that $$left|int_{gamma} exp(iz^2)dzright| leq frac{pibig(1-exp(-r^2)big)}{4r}$$ where $gamma(t)=re^{it}$, for $0leq t leq pi/4$ and $r > 0$.




I'm stuck. here is my attempt:



$|int_{gamma} e^{icdot z^2}dz|leq int_{gamma} |e^{icdot z^2}|dz=int_{gamma}|(e^{z^2})^i|dz$. As $t > 0$ on $0leq t leq pi/4$.
$Rightarrow|e^{r^2 e^{2it}}|=|e^{r^2}e^{e^{2it}}|>0$
$Rightarrowint_{0}^{pi/4} e^{r^2}e^{2it}rie^{it}dt=e^{r^2}riint_{0}^{pi/4} exp(e^{2it}+it)dt$

Let $alpha=e^{r^2}ri$
$Rightarrow alphaint_{0}^{pi/4}e^{cos2t}e^{icdot sin2t}(cos(t)+icdot sin(t)dt)$



Am I on track?






integration complex-analysis contour-integration integral-inequality holomorphic-functions







share|cite|improve this question









New contributor




Lincon Ribeiro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












Verify that $$left|int_{gamma} exp(iz^2)dzright| leq frac{pibig(1-exp(-r^2)big)}{4r}$$ where $gamma(t)=re^{it}$, for $0leq t leq pi/4$ and $r > 0$.




I'm stuck. here is my attempt:



$|int_{gamma} e^{icdot z^2}dz|leq int_{gamma} |e^{icdot z^2}|dz=int_{gamma}|(e^{z^2})^i|dz$. As $t > 0$ on $0leq t leq pi/4$.
$Rightarrow|e^{r^2 e^{2it}}|=|e^{r^2}e^{e^{2it}}|>0$
$Rightarrowint_{0}^{pi/4} e^{r^2}e^{2it}rie^{it}dt=e^{r^2}riint_{0}^{pi/4} exp(e^{2it}+it)dt$

Let $alpha=e^{r^2}ri$
$Rightarrow alphaint_{0}^{pi/4}e^{cos2t}e^{icdot sin2t}(cos(t)+icdot sin(t)dt)$



Am I on track?






integration complex-analysis contour-integration integral-inequality holomorphic-functions




integration complex-analysis contour-integration integral-inequality holomorphic-functions






share|cite|improve this question









New contributor




Lincon Ribeiro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Lincon Ribeiro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 15 hours ago





















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New contributor




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Lincon Ribeiro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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Check out our Code of Conduct.








  • 1




    What is $sen(t)$? Did you mean to write $sin(t)$?
    – LoveTooNap29
    yesterday










  • yeah, I corrected that. tks
    – Lincon Ribeiro
    yesterday






  • 1




    Hints: $|e^{i(x+iy)^2}|=e^{-2xy}$ and $sinphigeqslant 2phi/pi$ for $0leqslantphileqslantpi/2$.
    – metamorphy
    yesterday
















  • 1




    What is $sen(t)$? Did you mean to write $sin(t)$?
    – LoveTooNap29
    yesterday










  • yeah, I corrected that. tks
    – Lincon Ribeiro
    yesterday






  • 1




    Hints: $|e^{i(x+iy)^2}|=e^{-2xy}$ and $sinphigeqslant 2phi/pi$ for $0leqslantphileqslantpi/2$.
    – metamorphy
    yesterday










1




1




What is $sen(t)$? Did you mean to write $sin(t)$?
– LoveTooNap29
yesterday




What is $sen(t)$? Did you mean to write $sin(t)$?
– LoveTooNap29
yesterday












yeah, I corrected that. tks
– Lincon Ribeiro
yesterday




yeah, I corrected that. tks
– Lincon Ribeiro
yesterday




1




1




Hints: $|e^{i(x+iy)^2}|=e^{-2xy}$ and $sinphigeqslant 2phi/pi$ for $0leqslantphileqslantpi/2$.
– metamorphy
yesterday






Hints: $|e^{i(x+iy)^2}|=e^{-2xy}$ and $sinphigeqslant 2phi/pi$ for $0leqslantphileqslantpi/2$.
– metamorphy
yesterday












1 Answer
1






active

oldest

votes


















1














You have



$$ int_gamma e^{iz^2}dz = int_0^{pi/4} e^{ir^2(cos 2t + isin 2t)}ire^{it} dt $$



Taking the magnitude



$$ leftvert int_gamma e^{iz^2}dz rightvert le int_0^{pi/4} leftvert e^{ir^2(cos 2t + isin 2t)}ire^{it} rightvert dt = int_0^{pi/4} re^{-r^2sin 2t} dt $$



Since $sin 2t$ curves upwards on the interval $(0,pi/4)$, it always lies above its secant line from $t=0$ to $t=pi/4$, therefore



$$ sin 2t ge frac{4t}{pi}, quad forall t in left(0,frac{pi}{4}right) $$



And



$$ int_0^{pi/4} re^{-r^2sin 2t} dt le int_0^{pi/4} re^{-4r^2t/pi} dt = frac{pi}{4r}left(1 - e^{-r^2} right) $$






share|cite|improve this answer























  • @LinconRibeiro Can you remove your check mark? I made a mistake in my answer
    – Dylan
    15 hours ago










  • Ok, done. How did you get that term on the right side of the last inequality $frac {1-e^{-r^2}} {r^2}$? btw, I forgot to add that r must be greater than zero. I don't know if it changes anything.
    – Lincon Ribeiro
    15 hours ago






  • 1




    I updated my answer. It was simpler than I though.
    – Dylan
    14 hours ago











Your Answer





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1 Answer
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1














You have



$$ int_gamma e^{iz^2}dz = int_0^{pi/4} e^{ir^2(cos 2t + isin 2t)}ire^{it} dt $$



Taking the magnitude



$$ leftvert int_gamma e^{iz^2}dz rightvert le int_0^{pi/4} leftvert e^{ir^2(cos 2t + isin 2t)}ire^{it} rightvert dt = int_0^{pi/4} re^{-r^2sin 2t} dt $$



Since $sin 2t$ curves upwards on the interval $(0,pi/4)$, it always lies above its secant line from $t=0$ to $t=pi/4$, therefore



$$ sin 2t ge frac{4t}{pi}, quad forall t in left(0,frac{pi}{4}right) $$



And



$$ int_0^{pi/4} re^{-r^2sin 2t} dt le int_0^{pi/4} re^{-4r^2t/pi} dt = frac{pi}{4r}left(1 - e^{-r^2} right) $$






share|cite|improve this answer























  • @LinconRibeiro Can you remove your check mark? I made a mistake in my answer
    – Dylan
    15 hours ago










  • Ok, done. How did you get that term on the right side of the last inequality $frac {1-e^{-r^2}} {r^2}$? btw, I forgot to add that r must be greater than zero. I don't know if it changes anything.
    – Lincon Ribeiro
    15 hours ago






  • 1




    I updated my answer. It was simpler than I though.
    – Dylan
    14 hours ago
















1














You have



$$ int_gamma e^{iz^2}dz = int_0^{pi/4} e^{ir^2(cos 2t + isin 2t)}ire^{it} dt $$



Taking the magnitude



$$ leftvert int_gamma e^{iz^2}dz rightvert le int_0^{pi/4} leftvert e^{ir^2(cos 2t + isin 2t)}ire^{it} rightvert dt = int_0^{pi/4} re^{-r^2sin 2t} dt $$



Since $sin 2t$ curves upwards on the interval $(0,pi/4)$, it always lies above its secant line from $t=0$ to $t=pi/4$, therefore



$$ sin 2t ge frac{4t}{pi}, quad forall t in left(0,frac{pi}{4}right) $$



And



$$ int_0^{pi/4} re^{-r^2sin 2t} dt le int_0^{pi/4} re^{-4r^2t/pi} dt = frac{pi}{4r}left(1 - e^{-r^2} right) $$






share|cite|improve this answer























  • @LinconRibeiro Can you remove your check mark? I made a mistake in my answer
    – Dylan
    15 hours ago










  • Ok, done. How did you get that term on the right side of the last inequality $frac {1-e^{-r^2}} {r^2}$? btw, I forgot to add that r must be greater than zero. I don't know if it changes anything.
    – Lincon Ribeiro
    15 hours ago






  • 1




    I updated my answer. It was simpler than I though.
    – Dylan
    14 hours ago














1












1








1






You have



$$ int_gamma e^{iz^2}dz = int_0^{pi/4} e^{ir^2(cos 2t + isin 2t)}ire^{it} dt $$



Taking the magnitude



$$ leftvert int_gamma e^{iz^2}dz rightvert le int_0^{pi/4} leftvert e^{ir^2(cos 2t + isin 2t)}ire^{it} rightvert dt = int_0^{pi/4} re^{-r^2sin 2t} dt $$



Since $sin 2t$ curves upwards on the interval $(0,pi/4)$, it always lies above its secant line from $t=0$ to $t=pi/4$, therefore



$$ sin 2t ge frac{4t}{pi}, quad forall t in left(0,frac{pi}{4}right) $$



And



$$ int_0^{pi/4} re^{-r^2sin 2t} dt le int_0^{pi/4} re^{-4r^2t/pi} dt = frac{pi}{4r}left(1 - e^{-r^2} right) $$






share|cite|improve this answer














You have



$$ int_gamma e^{iz^2}dz = int_0^{pi/4} e^{ir^2(cos 2t + isin 2t)}ire^{it} dt $$



Taking the magnitude



$$ leftvert int_gamma e^{iz^2}dz rightvert le int_0^{pi/4} leftvert e^{ir^2(cos 2t + isin 2t)}ire^{it} rightvert dt = int_0^{pi/4} re^{-r^2sin 2t} dt $$



Since $sin 2t$ curves upwards on the interval $(0,pi/4)$, it always lies above its secant line from $t=0$ to $t=pi/4$, therefore



$$ sin 2t ge frac{4t}{pi}, quad forall t in left(0,frac{pi}{4}right) $$



And



$$ int_0^{pi/4} re^{-r^2sin 2t} dt le int_0^{pi/4} re^{-4r^2t/pi} dt = frac{pi}{4r}left(1 - e^{-r^2} right) $$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 14 hours ago

























answered 18 hours ago









Dylan

12.4k31026




12.4k31026












  • @LinconRibeiro Can you remove your check mark? I made a mistake in my answer
    – Dylan
    15 hours ago










  • Ok, done. How did you get that term on the right side of the last inequality $frac {1-e^{-r^2}} {r^2}$? btw, I forgot to add that r must be greater than zero. I don't know if it changes anything.
    – Lincon Ribeiro
    15 hours ago






  • 1




    I updated my answer. It was simpler than I though.
    – Dylan
    14 hours ago


















  • @LinconRibeiro Can you remove your check mark? I made a mistake in my answer
    – Dylan
    15 hours ago










  • Ok, done. How did you get that term on the right side of the last inequality $frac {1-e^{-r^2}} {r^2}$? btw, I forgot to add that r must be greater than zero. I don't know if it changes anything.
    – Lincon Ribeiro
    15 hours ago






  • 1




    I updated my answer. It was simpler than I though.
    – Dylan
    14 hours ago
















@LinconRibeiro Can you remove your check mark? I made a mistake in my answer
– Dylan
15 hours ago




@LinconRibeiro Can you remove your check mark? I made a mistake in my answer
– Dylan
15 hours ago












Ok, done. How did you get that term on the right side of the last inequality $frac {1-e^{-r^2}} {r^2}$? btw, I forgot to add that r must be greater than zero. I don't know if it changes anything.
– Lincon Ribeiro
15 hours ago




Ok, done. How did you get that term on the right side of the last inequality $frac {1-e^{-r^2}} {r^2}$? btw, I forgot to add that r must be greater than zero. I don't know if it changes anything.
– Lincon Ribeiro
15 hours ago




1




1




I updated my answer. It was simpler than I though.
– Dylan
14 hours ago




I updated my answer. It was simpler than I though.
– Dylan
14 hours ago










Lincon Ribeiro is a new contributor. Be nice, and check out our Code of Conduct.










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