How to find the Laplace transform of $\frac{1-\cos(t)}{t^2}$?

Question

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

Answer 1

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*}

Answer 2

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.

Answer 3

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

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

Answer 4

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