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$$ f(t)=\frac{1-\cos(t)}{t^2} $$ $$ F(S)= ? $$

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Did you read the answers below?:) –  B. S. Dec 18 '12 at 14:03
    
It was helpful. –  Amir Alizadeh Dec 22 '12 at 20:56
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5 Answers

up vote 13 down vote accepted
+50

Let

$$F(s) = \int_{0}^{\infty} f(t) \, e^{-st} \, dt = \int_{0}^{\infty} \frac{1-\cos t}{t^2} e^{-st} \, dt. $$

The function $f(t)$ satisfies the bound $ f(t) = O(1 \wedge t^{-2})$, thus it is absolutely integrable and we can apply Leibniz's integral to obtain

$$ F''(s) = \int_{0}^{\infty} (1-\cos t) \, e^{-st} \, dt = \frac{1}{s} - \frac{s}{s^2 + 1}. $$

Integrating and using the condition $F'(\infty) = 0$, we have

$$ F'(s) = \log s - \log \sqrt{s^2 + 1}. $$

Thus we have

$$F(s) = \int \left\{ \log s - \log \sqrt{s^2 + 1} \right\} \, ds. $$

The first term is easily integrated to yield $s \log s - s$. For the second term, note that

\begin{align*} -\int \log \sqrt{s^2 + 1} \, ds &= - s \log \sqrt{s^2 + 1} + \int \frac{s^2}{s^2 + 1} \, ds \\ &= - s \log \sqrt{s^2 + 1} + s - \arctan s + C. \end{align*}

Combining, we obtain

$$ F(s) = s \log s - s \log \sqrt{s^2 + 1} - \arctan s + C. $$

But since $F(\infty) = 0$, we must have $C = \frac{\pi}{2}$ and therefore

\begin{align*} F(s) &= s \log s - s \log \sqrt{s^2 + 1} - \arctan s + \frac{\pi}{2} \\ &= s \log \bigg( \frac{s}{\sqrt{s^2 + 1}} \bigg) + \arctan \left(\frac{1}{s}\right). \end{align*}

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Maple agrees with your answer. –  GEdgar Dec 15 '12 at 14:13
    
Nicely done. (+1) –  user26872 Dec 26 '12 at 8:14
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Besides Dennis's answer, using the following fact may be helpful:

If $~\mathcal L\left\{ f(t)\right\}=F(s)~$ and Laplace of the function $~g(t)=\dfrac{f(t)}{t}~$ exists, then

$$\mathcal L \left\{\dfrac{f(t)}{t}\right\}=\int\limits_s^{\infty}F(u)\,\mathrm du$$

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Thank you so much.It's helpful. –  Amir Alizadeh Dec 22 '12 at 20:58
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$$\begin{align*} \int\limits_s^\infty F(u)\,\mathrm du &=\int\limits_s^\infty\int\limits_0^\infty f(t)e^{-ut}\,\mathrm dt\,\mathrm du \\&=\int\limits_0^\infty f(t)\int\limits_s^\infty e^{-ut}\,\mathrm du\,\mathrm dt \\&=\int\limits_0^\infty f(t)\left.\frac{e^{-ut}}{-t}\right|_s^\infty \,\mathrm dt \\&=\int\limits_0^\infty \frac{f(t)}te^{-st}\,\mathrm dt \\&=\mathcal L\left\{\frac {f(t)}t \right\} \end{align*}$$ –  Elements in Space Dec 28 '12 at 8:57
    
@UnkleRhaukus: Thanks for the edit. ;-) –  B. S. Dec 28 '12 at 9:05
    
$~\ddot\smile~$ –  Elements in Space Dec 28 '12 at 9:16
    
$\color{blue}{\bf \ddot\smile \quad +1\quad}$ –  amWhy Apr 12 '13 at 0:09
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Probably the only way is to use the definition: $$F(s)=\int_0^\infty e^{-st}f(t) dt$$ But, at least according to WA, it has no elementary form.

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ Laplace Transform of $\ds{\fermi\pars{t} \equiv {1 - \cos\pars{t} \over t^{2}}:\ {\large ?}}$

\begin{align} &\color{#00f}{\large\int_{0}^{\infty}\fermi\pars{t}\expo{-st}\,\dd t} =\int_{0}^{\infty}\bracks{1 - \cos\pars{t}} \pars{\int_{0}^{\infty}x\expo{-tx}\,\dd x}\expo{-st}\,\dd t \\[3mm]&=\Re\int_{0}^{\infty}\dd x\,x\int_{0}^{\infty}\bracks{% \expo{-\pars{x + s}t} - \expo{-\pars{x + s - \ic}t}}\,\dd t =\Re\int_{0}^{\infty}x\pars{{1 \over x + s} - {1 \over x + s - \ic}}\,\dd x \\[3mm]&=\Re\int_{0}^{\infty}\pars{% -\,{s \over x + s} + {s - \ic \over x + s - \ic}}\,\dd x =\int_{0}^{\infty}\bracks{% -\,{s \over x + s} + {\pars{x + s}s + 1 \over \pars{x + s}^{2} + 1}}\,\dd x \\[3mm]&=\lim_{\Lambda \to \infty}\bracks{% \half\,s\ln\pars{\bracks{x + s}^{2} + 1 \over \bracks{x + s}^{2}} + \arctan\pars{x + s}}_{x = 0}^{x = \Lambda} \\[3mm]&=\color{#00f}{\large% -\,\half\,s\ln\pars{s^{2} + 1 \over s^{2}} + {\pi \over 2} - \arctan\pars{s}} \end{align}

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+1 for a nice solution. –  Integrals May 7 at 3:57
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As we know $$g(s)=\mathcal{L}\left\{1-\cos t\right\}=\mathcal{L}\left\{1\right\}-\mathcal{L}\left\{\cos t\right\}=\frac{1}{s}-\frac{s}{s^2+1}$$ Therefore, $$\mathcal{L}\left\{\frac{1-\cos t}{t}\right\}=\int_{s}^{\infty}g(\sigma)\, d\sigma=\int_{s}^{\infty} \frac{1}{\sigma}-\frac{\sigma}{\sigma^2+1}\, d\sigma$$ Since $$\int\frac{1}{\sigma}-\frac{\sigma}{\sigma^2+1}\, d\sigma=\ln \left|\sigma\right|-\frac{1}{2}\ln\left|\sigma^2+1\right|$$ we have that $$\mathcal{L}\left\{\frac{1-\cos t}{t}\right\}=\int_{s}^{\infty} \frac{1}{\sigma}-\frac{\sigma}{\sigma^2+1}\, d\sigma=1-\ln \left|\frac{s}{\sqrt{s^2+1}}\right|$$ Therefore, $$\mathcal{L}\left\{f\right\}=\int_{s}^{\infty}1-\ln \left|\frac{\sigma}{\sqrt{\sigma^2+1}}\right|\, d\sigma$$

I don't think you can go any further

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