Mixing
Verify that $left|int_{gamma} exp(iz^2)dzright| leq frac{pibig(1-exp(-r^2)big)}{4r}$ where...
$begingroup$
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
asked Jan 3 at 19:14
Lincon RibeiroLincon Ribeiro
506
$endgroup$
add a comment |
$begingroup$
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
asked Jan 3 at 19:14
Lincon RibeiroLincon Ribeiro
506
$endgroup$
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
add a comment |
$begingroup$
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
asked Jan 3 at 19:14
Lincon RibeiroLincon Ribeiro
506
$endgroup$
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
asked Jan 3 at 19:14
Lincon RibeiroLincon Ribeiro
506
asked Jan 3 at 19:14
Lincon RibeiroLincon Ribeiro
506
edited Jan 4 at 12:34
Lincon Ribeiro
asked Jan 3 at 19:14
Lincon RibeiroLincon Ribeiro
506
asked Jan 3 at 19:14
Lincon RibeiroLincon Ribeiro
506
asked Jan 3 at 19:14
Lincon RibeiroLincon Ribeiro
506
506
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
add a comment |
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
add a comment |
1 Answer
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$begingroup$
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
$$ int_0^{pi/4} re^{-r^2sin 2t} dt le int_0^{pi/4} re^{-4r^2t/pi} dt = frac{pi}{4r}left(1 - e^{-r^2} right) $$
answered Jan 4 at 9:02
DylanDylan
12.5k31026
$endgroup$
$begingroup$
@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
add a comment |
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1 Answer
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$begingroup$
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
$$ int_0^{pi/4} re^{-r^2sin 2t} dt le int_0^{pi/4} re^{-4r^2t/pi} dt = frac{pi}{4r}left(1 - e^{-r^2} right) $$
answered Jan 4 at 9:02
DylanDylan
12.5k31026
$endgroup$
$begingroup$
@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
add a comment |
$begingroup$
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
$$ int_0^{pi/4} re^{-r^2sin 2t} dt le int_0^{pi/4} re^{-4r^2t/pi} dt = frac{pi}{4r}left(1 - e^{-r^2} right) $$
answered Jan 4 at 9:02
DylanDylan
12.5k31026
$endgroup$
$begingroup$
@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
add a comment |
$begingroup$
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
$$ int_0^{pi/4} re^{-r^2sin 2t} dt le int_0^{pi/4} re^{-4r^2t/pi} dt = frac{pi}{4r}left(1 - e^{-r^2} right) $$
answered Jan 4 at 9:02
DylanDylan
12.5k31026
$endgroup$
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
$$ int_0^{pi/4} re^{-r^2sin 2t} dt le int_0^{pi/4} re^{-4r^2t/pi} dt = frac{pi}{4r}left(1 - e^{-r^2} right) $$
answered Jan 4 at 9:02
DylanDylan
12.5k31026
edited Jan 4 at 13:52
answered Jan 4 at 9:02
DylanDylan
12.5k31026
answered Jan 4 at 9:02
DylanDylan
12.5k31026
answered Jan 4 at 9:02
DylanDylan
12.5k31026
12.5k31026
$begingroup$
@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
add a comment |
$begingroup$
@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
add a comment |
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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