Inverse Laplace Transform of $F(s)=frac{1}{sqrt{s} coth(sqrt{s})-1}$.
$begingroup$
Dear All (best wishes for this new year 2019),
I try to find the inverse laplace transform of the function
$displaystyle F(s)=frac{1}{sqrt{s} coth(sqrt{s})-1}$. I check numerically that this function has no root in the right half complex plane and I used the folowing path. I found that the integral subpath $C_3+C_5$ is equal to $0$. However I am not able to compute the integrals on sub-path $C_2+C_6$ and $C_4$. Could you help me please. Thank you.
laplace-transform inverselaplace
edited Jan 2 at 22:53
El borito
589216
asked Jan 2 at 21:07
user425269user425269
415
$endgroup$
add a comment |
$begingroup$
Dear All (best wishes for this new year 2019),
I try to find the inverse laplace transform of the function
$displaystyle F(s)=frac{1}{sqrt{s} coth(sqrt{s})-1}$. I check numerically that this function has no root in the right half complex plane and I used the folowing path. I found that the integral subpath $C_3+C_5$ is equal to $0$. However I am not able to compute the integrals on sub-path $C_2+C_6$ and $C_4$. Could you help me please. Thank you.
laplace-transform inverselaplace
edited Jan 2 at 22:53
El borito
589216
asked Jan 2 at 21:07
user425269user425269
415
$endgroup$
$begingroup$
What information do you get from that fact that $F(s)$ has no root in the complex plane? $mathcal{L}^{-1} F(s)$ heavily depends on the singularities of $F(s)$, and there are plenty of them by the Picard little theorem.
$endgroup$
– Jack D'Aurizio
Jan 2 at 22:03
$begingroup$
On the other hand, since all the singularities of $F$ lie on the negative real axis and your simple contour $gamma$ does not enclose it, $oint_{gamma}F(s),ds=0$.
$endgroup$
– Jack D'Aurizio
Jan 2 at 22:07
add a comment |
$begingroup$
Dear All (best wishes for this new year 2019),
I try to find the inverse laplace transform of the function
$displaystyle F(s)=frac{1}{sqrt{s} coth(sqrt{s})-1}$. I check numerically that this function has no root in the right half complex plane and I used the folowing path. I found that the integral subpath $C_3+C_5$ is equal to $0$. However I am not able to compute the integrals on sub-path $C_2+C_6$ and $C_4$. Could you help me please. Thank you.
laplace-transform inverselaplace
edited Jan 2 at 22:53
El borito
589216
asked Jan 2 at 21:07
user425269user425269
415
$endgroup$
Dear All (best wishes for this new year 2019),
I try to find the inverse laplace transform of the function
$displaystyle F(s)=frac{1}{sqrt{s} coth(sqrt{s})-1}$. I check numerically that this function has no root in the right half complex plane and I used the folowing path. I found that the integral subpath $C_3+C_5$ is equal to $0$. However I am not able to compute the integrals on sub-path $C_2+C_6$ and $C_4$. Could you help me please. Thank you.
laplace-transform inverselaplace
laplace-transform inverselaplace
edited Jan 2 at 22:53
El borito
589216
asked Jan 2 at 21:07
user425269user425269
415
edited Jan 2 at 22:53
El borito
589216
asked Jan 2 at 21:07
user425269user425269
415
edited Jan 2 at 22:53
El borito
589216
edited Jan 2 at 22:53
El borito
589216
edited Jan 2 at 22:53
El borito
589216
589216
asked Jan 2 at 21:07
user425269user425269
415
asked Jan 2 at 21:07
user425269user425269
415
asked Jan 2 at 21:07
user425269user425269
415
415
$begingroup$
What information do you get from that fact that $F(s)$ has no root in the complex plane? $mathcal{L}^{-1} F(s)$ heavily depends on the singularities of $F(s)$, and there are plenty of them by the Picard little theorem.
$endgroup$
– Jack D'Aurizio
Jan 2 at 22:03
$begingroup$
On the other hand, since all the singularities of $F$ lie on the negative real axis and your simple contour $gamma$ does not enclose it, $oint_{gamma}F(s),ds=0$.
$endgroup$
– Jack D'Aurizio
Jan 2 at 22:07
add a comment |
$begingroup$
What information do you get from that fact that $F(s)$ has no root in the complex plane? $mathcal{L}^{-1} F(s)$ heavily depends on the singularities of $F(s)$, and there are plenty of them by the Picard little theorem.
$endgroup$
– Jack D'Aurizio
Jan 2 at 22:03
$begingroup$
On the other hand, since all the singularities of $F$ lie on the negative real axis and your simple contour $gamma$ does not enclose it, $oint_{gamma}F(s),ds=0$.
$endgroup$
– Jack D'Aurizio
Jan 2 at 22:07
$begingroup$
What information do you get from that fact that $F(s)$ has no root in the complex plane? $mathcal{L}^{-1} F(s)$ heavily depends on the singularities of $F(s)$, and there are plenty of them by the Picard little theorem.
$endgroup$
– Jack D'Aurizio
Jan 2 at 22:03
$begingroup$
What information do you get from that fact that $F(s)$ has no root in the complex plane? $mathcal{L}^{-1} F(s)$ heavily depends on the singularities of $F(s)$, and there are plenty of them by the Picard little theorem.
$endgroup$
– Jack D'Aurizio
Jan 2 at 22:03
$begingroup$
On the other hand, since all the singularities of $F$ lie on the negative real axis and your simple contour $gamma$ does not enclose it, $oint_{gamma}F(s),ds=0$.
$endgroup$
– Jack D'Aurizio
Jan 2 at 22:07
$begingroup$
On the other hand, since all the singularities of $F$ lie on the negative real axis and your simple contour $gamma$ does not enclose it, $oint_{gamma}F(s),ds=0$.
$endgroup$
– Jack D'Aurizio
Jan 2 at 22:07
add a comment |
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$begingroup$
What information do you get from that fact that $F(s)$ has no root in the complex plane? $mathcal{L}^{-1} F(s)$ heavily depends on the singularities of $F(s)$, and there are plenty of them by the Picard little theorem.
$endgroup$
– Jack D'Aurizio
Jan 2 at 22:03
$begingroup$
On the other hand, since all the singularities of $F$ lie on the negative real axis and your simple contour $gamma$ does not enclose it, $oint_{gamma}F(s),ds=0$.
$endgroup$
– Jack D'Aurizio
Jan 2 at 22:07