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












1












$begingroup$



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




    $begingroup$
    What is $sen(t)$? Did you mean to write $sin(t)$?
    $endgroup$
    – LoveTooNap29
    Jan 3 at 19:23










  • $begingroup$
    yeah, I corrected that. tks
    $endgroup$
    – Lincon Ribeiro
    Jan 3 at 19:25






  • 1




    $begingroup$
    Hints: $|e^{i(x+iy)^2}|=e^{-2xy}$ and $sinphigeqslant 2phi/pi$ for $0leqslantphileqslantpi/2$.
    $endgroup$
    – metamorphy
    Jan 4 at 2:29


















1












$begingroup$



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











$endgroup$








  • 1




    $begingroup$
    What is $sen(t)$? Did you mean to write $sin(t)$?
    $endgroup$
    – LoveTooNap29
    Jan 3 at 19:23










  • $begingroup$
    yeah, I corrected that. tks
    $endgroup$
    – Lincon Ribeiro
    Jan 3 at 19:25






  • 1




    $begingroup$
    Hints: $|e^{i(x+iy)^2}|=e^{-2xy}$ and $sinphigeqslant 2phi/pi$ for $0leqslantphileqslantpi/2$.
    $endgroup$
    – metamorphy
    Jan 4 at 2:29
















1












1








1


0



$begingroup$



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











$endgroup$





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















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 12:34







Lincon Ribeiro

















asked Jan 3 at 19:14









Lincon RibeiroLincon Ribeiro

617




617








  • 1




    $begingroup$
    What is $sen(t)$? Did you mean to write $sin(t)$?
    $endgroup$
    – LoveTooNap29
    Jan 3 at 19:23










  • $begingroup$
    yeah, I corrected that. tks
    $endgroup$
    – Lincon Ribeiro
    Jan 3 at 19:25






  • 1




    $begingroup$
    Hints: $|e^{i(x+iy)^2}|=e^{-2xy}$ and $sinphigeqslant 2phi/pi$ for $0leqslantphileqslantpi/2$.
    $endgroup$
    – metamorphy
    Jan 4 at 2:29
















  • 1




    $begingroup$
    What is $sen(t)$? Did you mean to write $sin(t)$?
    $endgroup$
    – LoveTooNap29
    Jan 3 at 19:23










  • $begingroup$
    yeah, I corrected that. tks
    $endgroup$
    – Lincon Ribeiro
    Jan 3 at 19:25






  • 1




    $begingroup$
    Hints: $|e^{i(x+iy)^2}|=e^{-2xy}$ and $sinphigeqslant 2phi/pi$ for $0leqslantphileqslantpi/2$.
    $endgroup$
    – metamorphy
    Jan 4 at 2:29










1




1




$begingroup$
What is $sen(t)$? Did you mean to write $sin(t)$?
$endgroup$
– LoveTooNap29
Jan 3 at 19:23




$begingroup$
What is $sen(t)$? Did you mean to write $sin(t)$?
$endgroup$
– LoveTooNap29
Jan 3 at 19:23












$begingroup$
yeah, I corrected that. tks
$endgroup$
– Lincon Ribeiro
Jan 3 at 19:25




$begingroup$
yeah, I corrected that. tks
$endgroup$
– Lincon Ribeiro
Jan 3 at 19:25




1




1




$begingroup$
Hints: $|e^{i(x+iy)^2}|=e^{-2xy}$ and $sinphigeqslant 2phi/pi$ for $0leqslantphileqslantpi/2$.
$endgroup$
– metamorphy
Jan 4 at 2:29






$begingroup$
Hints: $|e^{i(x+iy)^2}|=e^{-2xy}$ and $sinphigeqslant 2phi/pi$ for $0leqslantphileqslantpi/2$.
$endgroup$
– metamorphy
Jan 4 at 2:29












1 Answer
1






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oldest

votes


















1












$begingroup$

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











$endgroup$













  • $begingroup$
    @LinconRibeiro Can you remove your check mark? I made a mistake in my answer
    $endgroup$
    – Dylan
    Jan 4 at 12:17










  • $begingroup$
    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.
    $endgroup$
    – Lincon Ribeiro
    Jan 4 at 12:37






  • 1




    $begingroup$
    I updated my answer. It was simpler than I though.
    $endgroup$
    – Dylan
    Jan 4 at 13:47












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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

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1












$begingroup$

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











$endgroup$













  • $begingroup$
    @LinconRibeiro Can you remove your check mark? I made a mistake in my answer
    $endgroup$
    – Dylan
    Jan 4 at 12:17










  • $begingroup$
    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.
    $endgroup$
    – Lincon Ribeiro
    Jan 4 at 12:37






  • 1




    $begingroup$
    I updated my answer. It was simpler than I though.
    $endgroup$
    – Dylan
    Jan 4 at 13:47
















1












$begingroup$

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











$endgroup$













  • $begingroup$
    @LinconRibeiro Can you remove your check mark? I made a mistake in my answer
    $endgroup$
    – Dylan
    Jan 4 at 12:17










  • $begingroup$
    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.
    $endgroup$
    – Lincon Ribeiro
    Jan 4 at 12:37






  • 1




    $begingroup$
    I updated my answer. It was simpler than I though.
    $endgroup$
    – Dylan
    Jan 4 at 13:47














1












1








1





$begingroup$

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











$endgroup$



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 Jan 4 at 13:52

























answered Jan 4 at 9:02









DylanDylan

14.2k31127




14.2k31127












  • $begingroup$
    @LinconRibeiro Can you remove your check mark? I made a mistake in my answer
    $endgroup$
    – Dylan
    Jan 4 at 12:17










  • $begingroup$
    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.
    $endgroup$
    – Lincon Ribeiro
    Jan 4 at 12:37






  • 1




    $begingroup$
    I updated my answer. It was simpler than I though.
    $endgroup$
    – Dylan
    Jan 4 at 13:47


















  • $begingroup$
    @LinconRibeiro Can you remove your check mark? I made a mistake in my answer
    $endgroup$
    – Dylan
    Jan 4 at 12:17










  • $begingroup$
    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.
    $endgroup$
    – Lincon Ribeiro
    Jan 4 at 12:37






  • 1




    $begingroup$
    I updated my answer. It was simpler than I though.
    $endgroup$
    – Dylan
    Jan 4 at 13:47
















$begingroup$
@LinconRibeiro Can you remove your check mark? I made a mistake in my answer
$endgroup$
– Dylan
Jan 4 at 12:17




$begingroup$
@LinconRibeiro Can you remove your check mark? I made a mistake in my answer
$endgroup$
– Dylan
Jan 4 at 12:17












$begingroup$
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.
$endgroup$
– Lincon Ribeiro
Jan 4 at 12:37




$begingroup$
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.
$endgroup$
– Lincon Ribeiro
Jan 4 at 12:37




1




1




$begingroup$
I updated my answer. It was simpler than I though.
$endgroup$
– Dylan
Jan 4 at 13:47




$begingroup$
I updated my answer. It was simpler than I though.
$endgroup$
– Dylan
Jan 4 at 13:47


















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