Inverse Laplace Transform by Complex Inversion Theorem












1












$begingroup$


The questions asks to find $F(t) = L^{-1}{s^{-1/2}e^{-1/s}}$



I've made a branch cut along the negative real axis and produced a contour similar to this:



Integration along segments BE, LA go to 0 and since $f(s)$ has no singularities within the contour, the entire integration around it is 0 and we can find $F(t)$ as the integration along AB. So I have:



F(t) = $int_{AB}$ = - ($int_{EH}$ + $int_{HJK}$ + $int_{KL}$)



EH: Take $s=xe^{ipi}$



Then $$int_{EH} =int_{-R}^{-epsilon} frac{e^{st-1/s}}{sqrt s} ds =int_{R}^{epsilon} frac{ie^{1/x-xt}}{sqrt x} dx$$



Similarly for KL: Take $s=xe^{-ipi}$



Then $$int_{KL} =int_{-epsilon}^{-R} frac{e^{st-1/s}}{sqrt s} ds =int_{epsilon}^{R} frac{ie^{1/x-xt}}{sqrt x} dx$$



HJK: Take $s=epsilon e^{itheta}$



Then $$int_{HJK} =int_{pi}^{-pi} frac{e^{st-1/s}}{sqrt s} ds =int_{pi}^{-pi} frac{e^{epsilon t e^{itheta}-1/epsilon e^{itheta}}}{sqrt epsilon e^{itheta /2}} isqrt epsilon e^{itheta /2} ds =int_{pi}^{-pi} sqrt{epsilon} ie^{epsilon t e^{itheta} - (epsilon te^{itheta})^{-1}+ itheta/2} dx$$



Now, assuming these are correct (which might not be wise):



$$F(t) = -frac{1}{2pi i}lim_{Rto infty, epsilon to 0} int_{R}^{epsilon} frac{ie^{1/x-xt}}{sqrt x} dx + int_{epsilon}^{R} frac{ie^{1/x-xt}}{sqrt x} dx + int_{pi}^{-pi} sqrt{epsilon} ie^{epsilon t e^{itheta} - (epsilon te^{itheta})^{-1}+ itheta/2} dx$$



$$therefore F(t) = -frac{1}{2pi}lim_{Rto infty, epsilon to 0} int_{pi}^{-pi} sqrt{epsilon} e^{epsilon t e^{itheta} - (epsilon te^{itheta})^{-1}+ itheta/2} dx$$



I'm stuck here. Instructions were to let $epsilon to t^{-1/2}$ rather than 0 but I don't understand why and I can't evaluate this integral. I know the answer is $frac{cos(2sqrt t)}{sqrt{pi t}}$ from using a different method. Also how should I show the branch points analytically?






integration complex-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    The Maple code $$ with(inttrans): invlaplace(exp(-1/s)/s, s, t); $$ outputs $${{rm J}left(0,,2,sqrt {t}right)} .$$
    $endgroup$
    – user64494
    Oct 27 '13 at 7:48
















1












$begingroup$


The questions asks to find $F(t) = L^{-1}{s^{-1/2}e^{-1/s}}$



I've made a branch cut along the negative real axis and produced a contour similar to this:



Integration along segments BE, LA go to 0 and since $f(s)$ has no singularities within the contour, the entire integration around it is 0 and we can find $F(t)$ as the integration along AB. So I have:



F(t) = $int_{AB}$ = - ($int_{EH}$ + $int_{HJK}$ + $int_{KL}$)



EH: Take $s=xe^{ipi}$



Then $$int_{EH} =int_{-R}^{-epsilon} frac{e^{st-1/s}}{sqrt s} ds =int_{R}^{epsilon} frac{ie^{1/x-xt}}{sqrt x} dx$$



Similarly for KL: Take $s=xe^{-ipi}$



Then $$int_{KL} =int_{-epsilon}^{-R} frac{e^{st-1/s}}{sqrt s} ds =int_{epsilon}^{R} frac{ie^{1/x-xt}}{sqrt x} dx$$



HJK: Take $s=epsilon e^{itheta}$



Then $$int_{HJK} =int_{pi}^{-pi} frac{e^{st-1/s}}{sqrt s} ds =int_{pi}^{-pi} frac{e^{epsilon t e^{itheta}-1/epsilon e^{itheta}}}{sqrt epsilon e^{itheta /2}} isqrt epsilon e^{itheta /2} ds =int_{pi}^{-pi} sqrt{epsilon} ie^{epsilon t e^{itheta} - (epsilon te^{itheta})^{-1}+ itheta/2} dx$$



Now, assuming these are correct (which might not be wise):



$$F(t) = -frac{1}{2pi i}lim_{Rto infty, epsilon to 0} int_{R}^{epsilon} frac{ie^{1/x-xt}}{sqrt x} dx + int_{epsilon}^{R} frac{ie^{1/x-xt}}{sqrt x} dx + int_{pi}^{-pi} sqrt{epsilon} ie^{epsilon t e^{itheta} - (epsilon te^{itheta})^{-1}+ itheta/2} dx$$



$$therefore F(t) = -frac{1}{2pi}lim_{Rto infty, epsilon to 0} int_{pi}^{-pi} sqrt{epsilon} e^{epsilon t e^{itheta} - (epsilon te^{itheta})^{-1}+ itheta/2} dx$$



I'm stuck here. Instructions were to let $epsilon to t^{-1/2}$ rather than 0 but I don't understand why and I can't evaluate this integral. I know the answer is $frac{cos(2sqrt t)}{sqrt{pi t}}$ from using a different method. Also how should I show the branch points analytically?






integration complex-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    The Maple code $$ with(inttrans): invlaplace(exp(-1/s)/s, s, t); $$ outputs $${{rm J}left(0,,2,sqrt {t}right)} .$$
    $endgroup$
    – user64494
    Oct 27 '13 at 7:48














1












1








1





$begingroup$


The questions asks to find $F(t) = L^{-1}{s^{-1/2}e^{-1/s}}$



I've made a branch cut along the negative real axis and produced a contour similar to this:



Integration along segments BE, LA go to 0 and since $f(s)$ has no singularities within the contour, the entire integration around it is 0 and we can find $F(t)$ as the integration along AB. So I have:



F(t) = $int_{AB}$ = - ($int_{EH}$ + $int_{HJK}$ + $int_{KL}$)



EH: Take $s=xe^{ipi}$



Then $$int_{EH} =int_{-R}^{-epsilon} frac{e^{st-1/s}}{sqrt s} ds =int_{R}^{epsilon} frac{ie^{1/x-xt}}{sqrt x} dx$$



Similarly for KL: Take $s=xe^{-ipi}$



Then $$int_{KL} =int_{-epsilon}^{-R} frac{e^{st-1/s}}{sqrt s} ds =int_{epsilon}^{R} frac{ie^{1/x-xt}}{sqrt x} dx$$



HJK: Take $s=epsilon e^{itheta}$



Then $$int_{HJK} =int_{pi}^{-pi} frac{e^{st-1/s}}{sqrt s} ds =int_{pi}^{-pi} frac{e^{epsilon t e^{itheta}-1/epsilon e^{itheta}}}{sqrt epsilon e^{itheta /2}} isqrt epsilon e^{itheta /2} ds =int_{pi}^{-pi} sqrt{epsilon} ie^{epsilon t e^{itheta} - (epsilon te^{itheta})^{-1}+ itheta/2} dx$$



Now, assuming these are correct (which might not be wise):



$$F(t) = -frac{1}{2pi i}lim_{Rto infty, epsilon to 0} int_{R}^{epsilon} frac{ie^{1/x-xt}}{sqrt x} dx + int_{epsilon}^{R} frac{ie^{1/x-xt}}{sqrt x} dx + int_{pi}^{-pi} sqrt{epsilon} ie^{epsilon t e^{itheta} - (epsilon te^{itheta})^{-1}+ itheta/2} dx$$



$$therefore F(t) = -frac{1}{2pi}lim_{Rto infty, epsilon to 0} int_{pi}^{-pi} sqrt{epsilon} e^{epsilon t e^{itheta} - (epsilon te^{itheta})^{-1}+ itheta/2} dx$$



I'm stuck here. Instructions were to let $epsilon to t^{-1/2}$ rather than 0 but I don't understand why and I can't evaluate this integral. I know the answer is $frac{cos(2sqrt t)}{sqrt{pi t}}$ from using a different method. Also how should I show the branch points analytically?






integration complex-analysis







share|cite|improve this question











$endgroup$




The questions asks to find $F(t) = L^{-1}{s^{-1/2}e^{-1/s}}$



I've made a branch cut along the negative real axis and produced a contour similar to this:



Integration along segments BE, LA go to 0 and since $f(s)$ has no singularities within the contour, the entire integration around it is 0 and we can find $F(t)$ as the integration along AB. So I have:



F(t) = $int_{AB}$ = - ($int_{EH}$ + $int_{HJK}$ + $int_{KL}$)



EH: Take $s=xe^{ipi}$



Then $$int_{EH} =int_{-R}^{-epsilon} frac{e^{st-1/s}}{sqrt s} ds =int_{R}^{epsilon} frac{ie^{1/x-xt}}{sqrt x} dx$$



Similarly for KL: Take $s=xe^{-ipi}$



Then $$int_{KL} =int_{-epsilon}^{-R} frac{e^{st-1/s}}{sqrt s} ds =int_{epsilon}^{R} frac{ie^{1/x-xt}}{sqrt x} dx$$



HJK: Take $s=epsilon e^{itheta}$



Then $$int_{HJK} =int_{pi}^{-pi} frac{e^{st-1/s}}{sqrt s} ds =int_{pi}^{-pi} frac{e^{epsilon t e^{itheta}-1/epsilon e^{itheta}}}{sqrt epsilon e^{itheta /2}} isqrt epsilon e^{itheta /2} ds =int_{pi}^{-pi} sqrt{epsilon} ie^{epsilon t e^{itheta} - (epsilon te^{itheta})^{-1}+ itheta/2} dx$$



Now, assuming these are correct (which might not be wise):



$$F(t) = -frac{1}{2pi i}lim_{Rto infty, epsilon to 0} int_{R}^{epsilon} frac{ie^{1/x-xt}}{sqrt x} dx + int_{epsilon}^{R} frac{ie^{1/x-xt}}{sqrt x} dx + int_{pi}^{-pi} sqrt{epsilon} ie^{epsilon t e^{itheta} - (epsilon te^{itheta})^{-1}+ itheta/2} dx$$



$$therefore F(t) = -frac{1}{2pi}lim_{Rto infty, epsilon to 0} int_{pi}^{-pi} sqrt{epsilon} e^{epsilon t e^{itheta} - (epsilon te^{itheta})^{-1}+ itheta/2} dx$$



I'm stuck here. Instructions were to let $epsilon to t^{-1/2}$ rather than 0 but I don't understand why and I can't evaluate this integral. I know the answer is $frac{cos(2sqrt t)}{sqrt{pi t}}$ from using a different method. Also how should I show the branch points analytically?






integration complex-analysis




integration complex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 23 '18 at 21:18









Glorfindel

3,41981830




3,41981830










asked Oct 27 '13 at 7:43









GuestGuest

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  • $begingroup$
    The Maple code $$ with(inttrans): invlaplace(exp(-1/s)/s, s, t); $$ outputs $${{rm J}left(0,,2,sqrt {t}right)} .$$
    $endgroup$
    – user64494
    Oct 27 '13 at 7:48


















  • $begingroup$
    The Maple code $$ with(inttrans): invlaplace(exp(-1/s)/s, s, t); $$ outputs $${{rm J}left(0,,2,sqrt {t}right)} .$$
    $endgroup$
    – user64494
    Oct 27 '13 at 7:48
















$begingroup$
The Maple code $$ with(inttrans): invlaplace(exp(-1/s)/s, s, t); $$ outputs $${{rm J}left(0,,2,sqrt {t}right)} .$$
$endgroup$
– user64494
Oct 27 '13 at 7:48




$begingroup$
The Maple code $$ with(inttrans): invlaplace(exp(-1/s)/s, s, t); $$ outputs $${{rm J}left(0,,2,sqrt {t}right)} .$$
$endgroup$
– user64494
Oct 27 '13 at 7:48










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