 # Find the Laplace transformation (evaluating the improper integral that defines this transformation) of the real valued function f(t) of the real varia chillywilly12a 2021-03-04 Answered
Find the Laplace transformation (evaluating the improper integral that defines this transformation) of the real valued function f(t) of the real variable t>0. (Assume the parameter s appearing in the Laplace transformation, as a real variable).
$f\left(t\right)=2{t}^{2}-4\mathrm{cosh}\left(3t\right)+{e}^{{t}^{2}}$
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Step 1

here $f\left(t\right)=2{t}^{2}-4\mathrm{cosh}\left(3t\right)+{e}^{{t}^{2}}$
we know that
$L\left\{{t}^{n}\right\}=\frac{n!}{{s}^{n+1}},L\left\{\mathrm{cosh}\left(at\right)\right\}=\frac{s}{{s}^{2}-{a}^{2}}$
$L\left\{{e}^{{t}^{2}}\right\}=F\left(\frac{s}{2}\right)-\frac{1}{2}i\sqrt{\pi }\cdot {e}^{-\frac{{s}^{2}}{4}}$
Also , $L\left\{f\left(t\right)+g\left(t\right)+h\left(t\right)\right\}=L\left\{f\left(t\right)\right\}+L\left\{g\left(t\right)\right\}+L\left\{h\left(t\right)\right\}$
Taking Laplace transform of eq
$L\left\{f\left(t\right)\right\}=L\left\{2{t}^{2}-4\mathrm{cosh}\left(3t\right)+{e}^{{t}^{2}}\right\}$
$=2L\left\{{t}^{2}\right\}-4L\left\{\mathrm{cosh}\left(3t\right)\right\}+L\left\{{e}^{{t}^{2}}\right\}$
$=2\cdot \frac{2!}{{s}^{2+1}}-4\frac{s}{{s}^{2}-{3}^{2}}+F\left(\frac{s}{2}\right)-\frac{1}{2}i\sqrt{\pi }\cdot {e}^{-\frac{{s}^{2}}{4}}$
$L\left\{f\left(t\right)\right\}=\frac{4}{{s}^{3}}-\frac{4s}{{s}^{2}-9}+F\left(\frac{s}{2}\right)-\frac{1}{2}i\sqrt{\pi }\cdot {e}^{-\frac{{s}^{2}}{4}}$
Step 2
This is required Laplace transformation of given f(t).