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Question Number 220664 by SdC355 last updated on 17/May/25
example.. x(t)=∮_( C) ((s−2)/(s^2 −4s+6)) e^(st) ds Res_(s=2+(√2)i) {((s−2)/(s−2+(√2)i))}e^(st) +Res_(s=2−(√2)i) {((s−2)/(s−2−(√2)i))}e^(st) because Σ Res_(s=z_j ) {f(s)}e^(st) x(t)=(((√2)i)/(2(√2)i))e^(2t+(√2)it) +(((√2)i)/(2(√2)i))e^(2t−(√2)it) x(t)=e^(2t) (((e^((√2)it) +e^(−(√2)it) )/2)) ∴x(t)=e^(2t) cos((√2)t) Bromwich integral is defined as L_s ^(−1) {f(s)}= (1/(2πi)) ∮_( C) f(s)e^(st) ds if complex function f(s) is entire Does L_s ^(−1) {f(s)} dosen′t Exist? for example.... ∫_0 ^( ∞) J_ν (r)e^(−rt) dr=(((t+(√(t^2 +1)))^(−ν) )/( (√(t^2 +1)))) and we all know J_ν (s)=∮_( C) (((t+(√(t^2 +1)))^(−ν) )/( (√(t^2 +1)))) e^(st) dt But....can′t calculate...Σ_(h=1) ^N Res_(t=z_h ) {(((t+(√(t^2 +1)))^(−ν) )/( (√(t^2 +1))))} e^(st) and J_ν (s) can′t express by e^(λ_h t) , h=1,2,3... and by my Searching J_ν (s) defined as Σ_(j=0) ^∞ Res_(t=−j−(1/2)) {((𝚪(s+(1/2)ν))/(𝚪(1−s+(1/2)ν)))((s/2))^(2s) }
$$\:\mathrm{example}.. \\ $$$${x}\left({t}\right)=\oint_{\:{C}} \:\:\frac{{s}−\mathrm{2}}{{s}^{\mathrm{2}} −\mathrm{4}{s}+\mathrm{6}}\:{e}^{{st}} \:\mathrm{d}{s} \\ $$$$\mathrm{Res}_{{s}=\mathrm{2}+\sqrt{\mathrm{2}}\boldsymbol{{i}}} \left\{\frac{{s}−\mathrm{2}}{{s}−\mathrm{2}+\sqrt{\mathrm{2}}\boldsymbol{{i}}}\right\}{e}^{{st}} +\mathrm{Res}_{{s}=\mathrm{2}−\sqrt{\mathrm{2}}\boldsymbol{{i}}} \left\{\frac{{s}−\mathrm{2}}{{s}−\mathrm{2}−\sqrt{\mathrm{2}}\boldsymbol{{i}}}\right\}{e}^{{st}} \\ $$$$\mathrm{because}\:\Sigma\:\mathrm{Res}_{{s}={z}_{{j}} } \left\{{f}\left({s}\right)\right\}{e}^{{st}} \\ $$$${x}\left({t}\right)=\frac{\sqrt{\mathrm{2}}\boldsymbol{{i}}}{\mathrm{2}\sqrt{\mathrm{2}}\boldsymbol{{i}}}{e}^{\mathrm{2}{t}+\sqrt{\mathrm{2}}\boldsymbol{{i}}{t}} +\frac{\sqrt{\mathrm{2}}\boldsymbol{{i}}}{\mathrm{2}\sqrt{\mathrm{2}}\boldsymbol{{i}}}{e}^{\mathrm{2}{t}−\sqrt{\mathrm{2}}\boldsymbol{{i}}{t}} \: \\ $$$${x}\left({t}\right)={e}^{\mathrm{2}{t}} \left(\frac{{e}^{\sqrt{\mathrm{2}}\boldsymbol{{i}}{t}} +{e}^{−\sqrt{\mathrm{2}}\boldsymbol{{i}}{t}} }{\mathrm{2}}\right) \\ $$$$\therefore{x}\left({t}\right)={e}^{\mathrm{2}{t}} \mathrm{cos}\left(\sqrt{\mathrm{2}}{t}\right) \\ $$$$\mathrm{Bromwich}\:\mathrm{integral}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{as} \\ $$$$\mathcal{L}_{{s}} ^{−\mathrm{1}} \left\{{f}\left({s}\right)\right\}=\:\frac{\mathrm{1}}{\mathrm{2}\pi\boldsymbol{{i}}}\:\oint_{\:\mathrm{C}} \:{f}\left({s}\right){e}^{{st}} \:\mathrm{d}{s} \\ $$$$\mathrm{if}\:\mathrm{complex}\:\mathrm{function}\:\:{f}\left({s}\right)\:\mathrm{is}\:\mathrm{entire} \\ $$$$\mathrm{Does}\:\mathcal{L}_{{s}} ^{−\mathrm{1}} \left\{{f}\left({s}\right)\right\}\:\mathrm{dosen}'\mathrm{t}\:\mathrm{Exist}? \\ $$$$\mathrm{for}\:\mathrm{example}…. \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:{J}_{\nu} \left({r}\right){e}^{−{rt}} \mathrm{d}{r}=\frac{\left({t}+\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} }{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}\:\mathrm{and} \\ $$$$\mathrm{we}\:\mathrm{all}\:\mathrm{know}\:{J}_{\nu} \left({s}\right)=\oint_{\:\mathcal{C}} \:\:\frac{\left({t}+\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} }{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}\:{e}^{{st}} \:\mathrm{d}{t}\: \\ $$$$\mathrm{But}….\mathrm{can}'\mathrm{t}\:\mathrm{calculate}…\underset{{h}=\mathrm{1}} {\overset{{N}} {\sum}}\:\mathrm{Res}_{{t}={z}_{{h}} } \left\{\frac{\left({t}+\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} }{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}\right\}\:{e}^{{st}} \\ $$$$\mathrm{and}\:{J}_{\nu} \left({s}\right)\:\mathrm{can}'\mathrm{t}\:\mathrm{express}\:\mathrm{by}\:{e}^{\lambda_{{h}} {t}} \:,\:{h}=\mathrm{1},\mathrm{2},\mathrm{3}… \\ $$$$\mathrm{and}\:\mathrm{by}\:\mathrm{my}\:\mathrm{Searching} \\ $$$${J}_{\nu} \left({s}\right)\:\mathrm{defined}\:\mathrm{as}\:\underset{{j}=\mathrm{0}} {\overset{\infty} {\sum}}\:\mathrm{Res}_{{t}=−{j}−\frac{\mathrm{1}}{\mathrm{2}}} \left\{\frac{\boldsymbol{\Gamma}\left({s}+\frac{\mathrm{1}}{\mathrm{2}}\nu\right)}{\boldsymbol{\Gamma}\left(\mathrm{1}−{s}+\frac{\mathrm{1}}{\mathrm{2}}\nu\right)}\left(\frac{{s}}{\mathrm{2}}\right)^{\mathrm{2}{s}} \right\} \\ $$

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