# Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t)

Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by
$F\left(s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt$
where we assume s is a positive real number. For example, to find the Laplace transform of $f\left(t\right)={e}^{-t}$, the following improper integral is evaluated using integration by parts:
$F\left(s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}{e}^{-t}dt={\int }_{0}^{\mathrm{\infty }}{e}^{-\left(s+1\right)t}dt=\frac{1}{s+1}$
Verify the following Laplace transforms, where u is a real number.
$f\left(t\right)=\mathrm{cos}at\to F\left(s\right)=\frac{s}{{s}^{2}+{a}^{2}}$
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Step 1
We have given the function.
$f\left(t\right)=\mathrm{cos}at$
Step 2
To prove this we plug the values in the formula.
$L\le ft\left\{\mathrm{cos}atright\right\}\left(s\right)={\int }_{0}^{\to +\mathrm{\infty }}{e}^{-st}\mathrm{cos}atdt$
$=\underset{L\to \mathrm{\infty }}{lim}{\int }_{0}^{L}{e}^{-st}\mathrm{cos}atdt$
$=\underset{L\to \mathrm{\infty }}{lim}{\left[\frac{{e}^{-st}\left(-s\mathrm{cos}at+a\mathrm{sin}at\right)}{{\left(-s\right)}^{2}+{a}^{2}}\right]}_{0}^{L}$
$=\underset{L\to \mathrm{\infty }}{lim}\left(\frac{{e}^{-sL}\left(-s\mathrm{cos}aL+a\mathrm{sin}aL\right)}{{s}^{2}+{a}^{2}}-\frac{{e}^{-s×0}\left(-s\mathrm{cos}\left(0×a\right)+a\mathrm{sin}\left(0×a\right)\right)}{{s}^{2}+{a}^{2}}\right)$
$=\underset{L\to \mathrm{\infty }}{lim}\left(\frac{s\mathrm{cos}\left(0×a\right)-a\mathrm{sin}\left(0×a\right)}{{s}^{2}+{a}^{2}}-\frac{{e}^{-sL}\left(-s\mathrm{cos}aL+a\mathrm{sin}aL\right)}{{s}^{2}+{a}^{2}}\right)$
$=\frac{s\mathrm{cos}\left(0×a\right)-a\mathrm{sin}\left(0×a\right)}{{s}^{2}+{a}^{2}}-0$
$=\frac{s\mathrm{cos}0-a\mathrm{sin}0}{{s}^{2}+{a}^{2}}$
$=\frac{s}{{s}^{2}+{a}^{2}}$