Question

Find the inverse laplace transform of the arctangent of 1/p

\[\mathcal L^{-1}\{\arctan\tfrac1p\}\]

Answer 1

You have to find it?

Answer 2

yeah, i dont know how to start

Answer 3

I think working with the function's derivative might help.
\[\frac{d}{dp}\left[\arctan\frac{1}{p}\right]=-\frac{1}{p^2+1}\]

Answer 4

So in effect, you'd be finding the inverse transform of an integral of this function. Are there any rules you know about the transform of an integral?

Answer 5

something like this

Answer 6

sin(t)/t

Answer 7

\[\mathcal L^{-1}\{\arctan\tfrac1p\}=x(t)\\
-\frac1{(1/p)^2+1}={X(p)}\\
-\frac{p^2}{1+p^2}={X(p)}\\
\]

am i on the right track?

Answer 8

If \(\dfrac{\sin t}{t}\) is the answer, then using the rule is the right way to go.

Answer 9

\[L(\sin(t)=1/1+s2=d(\tan(p))/dp L(\sin(t)/t=\int\limits_{p}^{infinity}1/1+s2ds=\pi/2-\tan-1s=\cot-1s=\tan-1(1/s)\]

Answer 10

\[\newcommand \p \newcommand
\p \dd [1] {\,\mathrm d#1 }
\p \de [2] {\frac{\mathrm d #1}{\mathrm d#2} }
\p \Lin [1] {\operatorname{\mathcal L}^{-1}\left\{#1\right\} }
\p \intl [4] {\int\limits_{#1}^{#2}{#3}\dd{#4} }
\begin{align*}
\Lin{\arctan\tfrac1p}
&=\Lin{\intl p\infty{\de{}p(\arctan\tfrac1p)} p} \\
&=\Lin{\intl p\infty{\frac{-1/p^2}{1+(1/p)^2}}p} \\
&=\Lin{\intl p\infty{\frac{-1}{p^2+1}}p} \\
&=\frac{\sin t}t \\
\end{align*}\]

Answer 11

is this right ?

Answer 12

differential equations...I'm not even there yet...

Answer 13

i'm still not understanding this completely

Answer 14

Hows this? \[ \newcommand \p \newcommand
\p \Lap [1] {\operatorname{\mathcal L}\left\{\rule{0pt}{2.2ex}#1\right\} }
\p \dd [1] {\,\mathrm d#1 }
\p \de [2] {\frac{\mathrm d #1}{\mathrm d#2} }
\p \Lin [1] {\operatorname{\mathcal L}^{-1}\left\{#1\right\} }
\p \intl [4] {\int\limits_{#1}^{#2}{#3}\dd{#4} }
\begin{align*}
\Lin{\arctan\tfrac1p} &=\Lin{\intl p\infty{F(p)}p} =\frac{f(t)}t\\
\\
\arctan\tfrac1p &=\intl p\infty{F(p)}p \\
\de{}p\left(\arctan\tfrac1p\right) &=\de{}p\intl p\infty{F(p)}p \\
\frac1{1+1/p^2}\cdot-1/p^2 &=F(\infty)\cdot(\infty)'-F(p)\cdot(p)' \\
\frac{-1/p^2}{1+1/p^2} &=-F(p) \\
\frac{1}{p^2+1} &=F(p) \\
\\
\Lin{\frac{1}{p^2+1}} &=\Lin{F(p)} \\
\sin(t) &=f(t) \\
\\
\Lin{\arctan\tfrac1p} &=\frac{\sin t}t
\end{align*} \]

Answer 15

I'm not sure I do either, my class and textbook didn't cover Laplace past basic inverse transforms.