Laplace Transform Example: $mathcal{L}(e^{it}) = left[ dfrac{e^{(i - s)t}}{i - s} right]_0^infty = dfrac{1}{s...
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Laplace Transform Example: $mathcal{L}(e^{it}) = left[ dfrac{e^{(i - s)t}}{i - s} right]_0^infty = dfrac{1}{s - i}$?
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My textbook on Laplace transforms gives the following example:
$$begin{align} mathcal{L}(e^{it}) &= int_0^infty e^{-st} e^{it} dt \ &= int_0^infty e^{t(i - s)} dt \ &= left[ dfrac{e^{(i - s)t}}{i - s} right]_0^infty \ &= dfrac{1}{s - i} \ &= dfrac{s}{s^2 + 1} + idfrac{1}{s^2 + 1} end{align}$$
How did the author get from $left[ dfrac{e^{(i - s)t}}{i - s} right]_0^infty$ to $dfrac{1}{s - i}$? We actually get $left[ dfrac{e^{(i - s)t}}{i - s} right]_0^infty = dfrac{e^{(i - s)infty}}{i - s} - dfrac{1}{s - i}$, and I don't see how it's valid to just discard $dfrac{e^{(i - s)infty}}{i - s}$.
I would greatly appreciate it if people could please take the time to explain what's going on here.
real-analysis integration complex-analysis laplace-transform
asked Mar 21 at 4:39
The PointerThe Pointer
2,48731641
$endgroup$
add a comment |
$begingroup$
My textbook on Laplace transforms gives the following example:
$$begin{align} mathcal{L}(e^{it}) &= int_0^infty e^{-st} e^{it} dt \ &= int_0^infty e^{t(i - s)} dt \ &= left[ dfrac{e^{(i - s)t}}{i - s} right]_0^infty \ &= dfrac{1}{s - i} \ &= dfrac{s}{s^2 + 1} + idfrac{1}{s^2 + 1} end{align}$$
How did the author get from $left[ dfrac{e^{(i - s)t}}{i - s} right]_0^infty$ to $dfrac{1}{s - i}$? We actually get $left[ dfrac{e^{(i - s)t}}{i - s} right]_0^infty = dfrac{e^{(i - s)infty}}{i - s} - dfrac{1}{s - i}$, and I don't see how it's valid to just discard $dfrac{e^{(i - s)infty}}{i - s}$.
I would greatly appreciate it if people could please take the time to explain what's going on here.
real-analysis integration complex-analysis laplace-transform
asked Mar 21 at 4:39
The PointerThe Pointer
2,48731641
$endgroup$
add a comment |
$begingroup$
My textbook on Laplace transforms gives the following example:
$$begin{align} mathcal{L}(e^{it}) &= int_0^infty e^{-st} e^{it} dt \ &= int_0^infty e^{t(i - s)} dt \ &= left[ dfrac{e^{(i - s)t}}{i - s} right]_0^infty \ &= dfrac{1}{s - i} \ &= dfrac{s}{s^2 + 1} + idfrac{1}{s^2 + 1} end{align}$$
How did the author get from $left[ dfrac{e^{(i - s)t}}{i - s} right]_0^infty$ to $dfrac{1}{s - i}$? We actually get $left[ dfrac{e^{(i - s)t}}{i - s} right]_0^infty = dfrac{e^{(i - s)infty}}{i - s} - dfrac{1}{s - i}$, and I don't see how it's valid to just discard $dfrac{e^{(i - s)infty}}{i - s}$.
I would greatly appreciate it if people could please take the time to explain what's going on here.
real-analysis integration complex-analysis laplace-transform
asked Mar 21 at 4:39
The PointerThe Pointer
2,48731641
$endgroup$
My textbook on Laplace transforms gives the following example:
$$begin{align} mathcal{L}(e^{it}) &= int_0^infty e^{-st} e^{it} dt \ &= int_0^infty e^{t(i - s)} dt \ &= left[ dfrac{e^{(i - s)t}}{i - s} right]_0^infty \ &= dfrac{1}{s - i} \ &= dfrac{s}{s^2 + 1} + idfrac{1}{s^2 + 1} end{align}$$
How did the author get from $left[ dfrac{e^{(i - s)t}}{i - s} right]_0^infty$ to $dfrac{1}{s - i}$? We actually get $left[ dfrac{e^{(i - s)t}}{i - s} right]_0^infty = dfrac{e^{(i - s)infty}}{i - s} - dfrac{1}{s - i}$, and I don't see how it's valid to just discard $dfrac{e^{(i - s)infty}}{i - s}$.
I would greatly appreciate it if people could please take the time to explain what's going on here.
real-analysis integration complex-analysis laplace-transform
real-analysis integration complex-analysis laplace-transform
asked Mar 21 at 4:39
The PointerThe Pointer
2,48731641
asked Mar 21 at 4:39
The PointerThe Pointer
2,48731641
asked Mar 21 at 4:39
The PointerThe Pointer
2,48731641
asked Mar 21 at 4:39
The PointerThe Pointer
2,48731641
asked Mar 21 at 4:39
The PointerThe Pointer
2,48731641
2,48731641
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that De Moivre's formula tells you that $e^{it}$ for any $tin mathbb{R}$ is bounded because $sin t$ and $cos t$ are bounded.
$$begin{aligned}mathcal{L(e^{it})}&=int_{0}^{infty}e^{(i-s)t}mathrm dt\&=left[dfrac{e^{(i-s)t}}{i-s}right]_{0}^{infty}\&=underbrace{lim_{tto infty}dfrac{e^{it}}{e^{st}(i-s)}}_{to 0}-dfrac{1}{i-s}\&to dfrac{1}{s-i}end{aligned}$$
answered Mar 21 at 4:44
Paras KhoslaParas Khosla
3,051625
$endgroup$
$begingroup$
But it’s not clear why that should be the case, since what you actually have is an indeterminate form $dfrac{infty}{infty}$. It seems to me like some deeper level of justification is required here by using concepts from complex analysis?
$endgroup$
– The Pointer
Mar 21 at 4:46
1
$begingroup$
De Moivre's formula tells you that $e^{it}=cos t+isin t$ and sine and cosine are bounded for $tin mathbb{R}$ so clearly $e^{st}$ outspeeds $e^{it}$ in growth as $t to infty$.
$endgroup$
– Paras Khosla
Mar 21 at 4:48
$begingroup$
yes, you are right. This explains it very well. Thank you for the clarification!
$endgroup$
– The Pointer
Mar 21 at 4:51
$begingroup$
You're welcome @ThePointer Cheers :))
$endgroup$
– Paras Khosla
Mar 21 at 4:52
add a comment |
$begingroup$
$newcommand{Re}{operatorname{Re}}$Note that for any real $t$ and complex $s$, we have $$left|e^{(i-s)t}right| = e^{Re((i-s)t)} = e^{-at},$$ where $a=Re(s)$. (Recall in general that $left|e^zright| = e^{Re(z)}$ for any complex $z$.) So you can see that if $a=Re(s) > 0$, we have $limlimits_{ttoinfty}left|e^{(i-s)t}right| = 0$, and so $$limlimits_{ttoinfty}frac{e^{(i-s)t}}{i-s}=0.$$
answered Mar 21 at 4:51
Minus One-TwelfthMinus One-Twelfth
3,363413
$endgroup$
$begingroup$
This is interesting. Where is the fact that $left|e^zright| = e^{Re(z)}$ from?
$endgroup$
– The Pointer
Mar 21 at 5:19
1
$begingroup$
This can be found by writing $z=x+iy $ where $x$ and $y$ are the real and imaginary parts of $z$ respectively. Then $e^z = e^{x+iy} = e^x e^{iy}$. Hence $$begin{align}left|e^zright| &= left|e^xright|left|e^{iy}right| \ &= e^x cdot 1 \ &= e^x \ &= e^{operatorname{Re}(z)}.end{align}$$
$endgroup$
– Minus One-Twelfth
Mar 21 at 5:22
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that De Moivre's formula tells you that $e^{it}$ for any $tin mathbb{R}$ is bounded because $sin t$ and $cos t$ are bounded.
$$begin{aligned}mathcal{L(e^{it})}&=int_{0}^{infty}e^{(i-s)t}mathrm dt\&=left[dfrac{e^{(i-s)t}}{i-s}right]_{0}^{infty}\&=underbrace{lim_{tto infty}dfrac{e^{it}}{e^{st}(i-s)}}_{to 0}-dfrac{1}{i-s}\&to dfrac{1}{s-i}end{aligned}$$
answered Mar 21 at 4:44
Paras KhoslaParas Khosla
3,051625
$endgroup$
$begingroup$
But it’s not clear why that should be the case, since what you actually have is an indeterminate form $dfrac{infty}{infty}$. It seems to me like some deeper level of justification is required here by using concepts from complex analysis?
$endgroup$
– The Pointer
Mar 21 at 4:46
1
$begingroup$
De Moivre's formula tells you that $e^{it}=cos t+isin t$ and sine and cosine are bounded for $tin mathbb{R}$ so clearly $e^{st}$ outspeeds $e^{it}$ in growth as $t to infty$.
$endgroup$
– Paras Khosla
Mar 21 at 4:48
$begingroup$
yes, you are right. This explains it very well. Thank you for the clarification!
$endgroup$
– The Pointer
Mar 21 at 4:51
$begingroup$
You're welcome @ThePointer Cheers :))
$endgroup$
– Paras Khosla
Mar 21 at 4:52
add a comment |
$begingroup$
Note that De Moivre's formula tells you that $e^{it}$ for any $tin mathbb{R}$ is bounded because $sin t$ and $cos t$ are bounded.
$$begin{aligned}mathcal{L(e^{it})}&=int_{0}^{infty}e^{(i-s)t}mathrm dt\&=left[dfrac{e^{(i-s)t}}{i-s}right]_{0}^{infty}\&=underbrace{lim_{tto infty}dfrac{e^{it}}{e^{st}(i-s)}}_{to 0}-dfrac{1}{i-s}\&to dfrac{1}{s-i}end{aligned}$$
answered Mar 21 at 4:44
Paras KhoslaParas Khosla
3,051625
$endgroup$
$begingroup$
But it’s not clear why that should be the case, since what you actually have is an indeterminate form $dfrac{infty}{infty}$. It seems to me like some deeper level of justification is required here by using concepts from complex analysis?
$endgroup$
– The Pointer
Mar 21 at 4:46
1
$begingroup$
De Moivre's formula tells you that $e^{it}=cos t+isin t$ and sine and cosine are bounded for $tin mathbb{R}$ so clearly $e^{st}$ outspeeds $e^{it}$ in growth as $t to infty$.
$endgroup$
– Paras Khosla
Mar 21 at 4:48
$begingroup$
yes, you are right. This explains it very well. Thank you for the clarification!
$endgroup$
– The Pointer
Mar 21 at 4:51
$begingroup$
You're welcome @ThePointer Cheers :))
$endgroup$
– Paras Khosla
Mar 21 at 4:52
add a comment |
$begingroup$
Note that De Moivre's formula tells you that $e^{it}$ for any $tin mathbb{R}$ is bounded because $sin t$ and $cos t$ are bounded.
$$begin{aligned}mathcal{L(e^{it})}&=int_{0}^{infty}e^{(i-s)t}mathrm dt\&=left[dfrac{e^{(i-s)t}}{i-s}right]_{0}^{infty}\&=underbrace{lim_{tto infty}dfrac{e^{it}}{e^{st}(i-s)}}_{to 0}-dfrac{1}{i-s}\&to dfrac{1}{s-i}end{aligned}$$
answered Mar 21 at 4:44
Paras KhoslaParas Khosla
3,051625
$endgroup$
Note that De Moivre's formula tells you that $e^{it}$ for any $tin mathbb{R}$ is bounded because $sin t$ and $cos t$ are bounded.
$$begin{aligned}mathcal{L(e^{it})}&=int_{0}^{infty}e^{(i-s)t}mathrm dt\&=left[dfrac{e^{(i-s)t}}{i-s}right]_{0}^{infty}\&=underbrace{lim_{tto infty}dfrac{e^{it}}{e^{st}(i-s)}}_{to 0}-dfrac{1}{i-s}\&to dfrac{1}{s-i}end{aligned}$$
answered Mar 21 at 4:44
Paras KhoslaParas Khosla
3,051625
edited Mar 21 at 4:46
answered Mar 21 at 4:44
Paras KhoslaParas Khosla
3,051625
answered Mar 21 at 4:44
Paras KhoslaParas Khosla
3,051625
answered Mar 21 at 4:44
Paras KhoslaParas Khosla
3,051625
3,051625
$begingroup$
But it’s not clear why that should be the case, since what you actually have is an indeterminate form $dfrac{infty}{infty}$. It seems to me like some deeper level of justification is required here by using concepts from complex analysis?
$endgroup$
– The Pointer
Mar 21 at 4:46
1
$begingroup$
De Moivre's formula tells you that $e^{it}=cos t+isin t$ and sine and cosine are bounded for $tin mathbb{R}$ so clearly $e^{st}$ outspeeds $e^{it}$ in growth as $t to infty$.
$endgroup$
– Paras Khosla
Mar 21 at 4:48
$begingroup$
yes, you are right. This explains it very well. Thank you for the clarification!
$endgroup$
– The Pointer
Mar 21 at 4:51
$begingroup$
You're welcome @ThePointer Cheers :))
$endgroup$
– Paras Khosla
Mar 21 at 4:52
add a comment |
$begingroup$
But it’s not clear why that should be the case, since what you actually have is an indeterminate form $dfrac{infty}{infty}$. It seems to me like some deeper level of justification is required here by using concepts from complex analysis?
$endgroup$
– The Pointer
Mar 21 at 4:46
1
$begingroup$
De Moivre's formula tells you that $e^{it}=cos t+isin t$ and sine and cosine are bounded for $tin mathbb{R}$ so clearly $e^{st}$ outspeeds $e^{it}$ in growth as $t to infty$.
$endgroup$
– Paras Khosla
Mar 21 at 4:48
$begingroup$
yes, you are right. This explains it very well. Thank you for the clarification!
$endgroup$
– The Pointer
Mar 21 at 4:51
$begingroup$
You're welcome @ThePointer Cheers :))
$endgroup$
– Paras Khosla
Mar 21 at 4:52
$begingroup$
But it’s not clear why that should be the case, since what you actually have is an indeterminate form $dfrac{infty}{infty}$. It seems to me like some deeper level of justification is required here by using concepts from complex analysis?
$endgroup$
– The Pointer
Mar 21 at 4:46
$begingroup$
But it’s not clear why that should be the case, since what you actually have is an indeterminate form $dfrac{infty}{infty}$. It seems to me like some deeper level of justification is required here by using concepts from complex analysis?
$endgroup$
– The Pointer
Mar 21 at 4:46
1
1
$begingroup$
De Moivre's formula tells you that $e^{it}=cos t+isin t$ and sine and cosine are bounded for $tin mathbb{R}$ so clearly $e^{st}$ outspeeds $e^{it}$ in growth as $t to infty$.
$endgroup$
– Paras Khosla
Mar 21 at 4:48
$begingroup$
De Moivre's formula tells you that $e^{it}=cos t+isin t$ and sine and cosine are bounded for $tin mathbb{R}$ so clearly $e^{st}$ outspeeds $e^{it}$ in growth as $t to infty$.
$endgroup$
– Paras Khosla
Mar 21 at 4:48
$begingroup$
yes, you are right. This explains it very well. Thank you for the clarification!
$endgroup$
– The Pointer
Mar 21 at 4:51
$begingroup$
yes, you are right. This explains it very well. Thank you for the clarification!
$endgroup$
– The Pointer
Mar 21 at 4:51
$begingroup$
You're welcome @ThePointer Cheers :))
$endgroup$
– Paras Khosla
Mar 21 at 4:52
$begingroup$
You're welcome @ThePointer Cheers :))
$endgroup$
– Paras Khosla
Mar 21 at 4:52
add a comment |
$begingroup$
$newcommand{Re}{operatorname{Re}}$Note that for any real $t$ and complex $s$, we have $$left|e^{(i-s)t}right| = e^{Re((i-s)t)} = e^{-at},$$ where $a=Re(s)$. (Recall in general that $left|e^zright| = e^{Re(z)}$ for any complex $z$.) So you can see that if $a=Re(s) > 0$, we have $limlimits_{ttoinfty}left|e^{(i-s)t}right| = 0$, and so $$limlimits_{ttoinfty}frac{e^{(i-s)t}}{i-s}=0.$$
answered Mar 21 at 4:51
Minus One-TwelfthMinus One-Twelfth
3,363413
$endgroup$
$begingroup$
This is interesting. Where is the fact that $left|e^zright| = e^{Re(z)}$ from?
$endgroup$
– The Pointer
Mar 21 at 5:19
1
$begingroup$
This can be found by writing $z=x+iy $ where $x$ and $y$ are the real and imaginary parts of $z$ respectively. Then $e^z = e^{x+iy} = e^x e^{iy}$. Hence $$begin{align}left|e^zright| &= left|e^xright|left|e^{iy}right| \ &= e^x cdot 1 \ &= e^x \ &= e^{operatorname{Re}(z)}.end{align}$$
$endgroup$
– Minus One-Twelfth
Mar 21 at 5:22
add a comment |
$begingroup$
$newcommand{Re}{operatorname{Re}}$Note that for any real $t$ and complex $s$, we have $$left|e^{(i-s)t}right| = e^{Re((i-s)t)} = e^{-at},$$ where $a=Re(s)$. (Recall in general that $left|e^zright| = e^{Re(z)}$ for any complex $z$.) So you can see that if $a=Re(s) > 0$, we have $limlimits_{ttoinfty}left|e^{(i-s)t}right| = 0$, and so $$limlimits_{ttoinfty}frac{e^{(i-s)t}}{i-s}=0.$$
answered Mar 21 at 4:51
Minus One-TwelfthMinus One-Twelfth
3,363413
$endgroup$
$begingroup$
This is interesting. Where is the fact that $left|e^zright| = e^{Re(z)}$ from?
$endgroup$
– The Pointer
Mar 21 at 5:19
1
$begingroup$
This can be found by writing $z=x+iy $ where $x$ and $y$ are the real and imaginary parts of $z$ respectively. Then $e^z = e^{x+iy} = e^x e^{iy}$. Hence $$begin{align}left|e^zright| &= left|e^xright|left|e^{iy}right| \ &= e^x cdot 1 \ &= e^x \ &= e^{operatorname{Re}(z)}.end{align}$$
$endgroup$
– Minus One-Twelfth
Mar 21 at 5:22
add a comment |
$begingroup$
$newcommand{Re}{operatorname{Re}}$Note that for any real $t$ and complex $s$, we have $$left|e^{(i-s)t}right| = e^{Re((i-s)t)} = e^{-at},$$ where $a=Re(s)$. (Recall in general that $left|e^zright| = e^{Re(z)}$ for any complex $z$.) So you can see that if $a=Re(s) > 0$, we have $limlimits_{ttoinfty}left|e^{(i-s)t}right| = 0$, and so $$limlimits_{ttoinfty}frac{e^{(i-s)t}}{i-s}=0.$$
answered Mar 21 at 4:51
Minus One-TwelfthMinus One-Twelfth
3,363413
$endgroup$
$newcommand{Re}{operatorname{Re}}$Note that for any real $t$ and complex $s$, we have $$left|e^{(i-s)t}right| = e^{Re((i-s)t)} = e^{-at},$$ where $a=Re(s)$. (Recall in general that $left|e^zright| = e^{Re(z)}$ for any complex $z$.) So you can see that if $a=Re(s) > 0$, we have $limlimits_{ttoinfty}left|e^{(i-s)t}right| = 0$, and so $$limlimits_{ttoinfty}frac{e^{(i-s)t}}{i-s}=0.$$
answered Mar 21 at 4:51
Minus One-TwelfthMinus One-Twelfth
3,363413
answered Mar 21 at 4:51
Minus One-TwelfthMinus One-Twelfth
3,363413
answered Mar 21 at 4:51
Minus One-TwelfthMinus One-Twelfth
3,363413
answered Mar 21 at 4:51
Minus One-TwelfthMinus One-Twelfth
3,363413
3,363413
$begingroup$
This is interesting. Where is the fact that $left|e^zright| = e^{Re(z)}$ from?
$endgroup$
– The Pointer
Mar 21 at 5:19
1
$begingroup$
This can be found by writing $z=x+iy $ where $x$ and $y$ are the real and imaginary parts of $z$ respectively. Then $e^z = e^{x+iy} = e^x e^{iy}$. Hence $$begin{align}left|e^zright| &= left|e^xright|left|e^{iy}right| \ &= e^x cdot 1 \ &= e^x \ &= e^{operatorname{Re}(z)}.end{align}$$
$endgroup$
– Minus One-Twelfth
Mar 21 at 5:22
add a comment |
$begingroup$
This is interesting. Where is the fact that $left|e^zright| = e^{Re(z)}$ from?
$endgroup$
– The Pointer
Mar 21 at 5:19
1
$begingroup$
This can be found by writing $z=x+iy $ where $x$ and $y$ are the real and imaginary parts of $z$ respectively. Then $e^z = e^{x+iy} = e^x e^{iy}$. Hence $$begin{align}left|e^zright| &= left|e^xright|left|e^{iy}right| \ &= e^x cdot 1 \ &= e^x \ &= e^{operatorname{Re}(z)}.end{align}$$
$endgroup$
– Minus One-Twelfth
Mar 21 at 5:22
$begingroup$
This is interesting. Where is the fact that $left|e^zright| = e^{Re(z)}$ from?
$endgroup$
– The Pointer
Mar 21 at 5:19
$begingroup$
This is interesting. Where is the fact that $left|e^zright| = e^{Re(z)}$ from?
$endgroup$
– The Pointer
Mar 21 at 5:19
1
1
$begingroup$
This can be found by writing $z=x+iy $ where $x$ and $y$ are the real and imaginary parts of $z$ respectively. Then $e^z = e^{x+iy} = e^x e^{iy}$. Hence $$begin{align}left|e^zright| &= left|e^xright|left|e^{iy}right| \ &= e^x cdot 1 \ &= e^x \ &= e^{operatorname{Re}(z)}.end{align}$$
$endgroup$
– Minus One-Twelfth
Mar 21 at 5:22
$begingroup$
This can be found by writing $z=x+iy $ where $x$ and $y$ are the real and imaginary parts of $z$ respectively. Then $e^z = e^{x+iy} = e^x e^{iy}$. Hence $$begin{align}left|e^zright| &= left|e^xright|left|e^{iy}right| \ &= e^x cdot 1 \ &= e^x \ &= e^{operatorname{Re}(z)}.end{align}$$
$endgroup$
– Minus One-Twelfth
Mar 21 at 5:22
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