How to solve $\mathcal{L}\left\{\frac{\sin(at)}{t^2}\right\}$ and $\mathcal{L}\left\{\frac{\cos(at)}{t}\right\}$

Question

Background : While reviewing my old answer from here, where I had solved this integral,

$$\int_0^\infty\frac{x}{(e^x-1)(x^2+4\pi^2)}dx$$

I found this question and I'm trying to solve this,

$$I=\int_0^\infty\frac{x}{(e^x-1)(x^2+4\pi^2)^2}dx$$

$$\color{red}{\mathcal{L}\left(\frac{\sin(a t)-a t \cos(a t) }{2a^2}\right)=\frac{a}{(s^2+a^2)^2}\underset{s=2\pi, a=x}\implies \mathcal{L}\left(\frac{\sin(x t)-x t \cos(x t) }{2x^2}\right)=\frac{x}{(x^2+4\pi^2)^2}}$$

$$\color{red}{\mathcal{L}\left(\frac{\sin(x t)-x t \cos(x t) }{2x^2}\right)=\frac{x}{(x^2+4\pi^2)^2}\implies\int_0^\infty e^{-st} \left(\frac{\sin(x t)-x t \cos(x t) }{2x^2}\right)\,dt}$$

Therefore,

$$I=\frac12\int_0^\infty\frac{1}{e^x-1}\int_0^\infty e^{-2\pi t} \left(\frac{\sin(x t)-x t \cos(x t) }{x^2}\right)\,dt\,dx$$

$$2I=\int_0^\infty\frac{1}{e^x-1}\left(\int_0^\infty e^{-2\pi t} \frac{\sin(xt)}{x^2}\,dt-\int_0^\infty e^{-2\pi t} \frac{t\cos(xt)}{x}\,dt\right)\,dx$$

$$2I=\int_0^\infty e^{-2\pi t} \int_0^\infty\frac{1}{e^x-1}\frac{\sin(xt)}{x^2}\,dx\,dt-\int_0^\infty te^{-2\pi t} \int_0^\infty\frac{1}{e^x-1}\frac{\cos(xt)}{x}\,dx\,dt$$

$$\color{red}{\sum_{n=1}^\infty e^{-nx}=\frac{1}{e^x-1}}$$

$$2I=\int_0^\infty e^{-2\pi t} \sum_{n=1}^\infty\int_0^\infty e^{-nx}\frac{\sin(xt)}{x^2}\,dx\,dt-\int_0^\infty te^{-2\pi t} \sum_{n=1}^\infty\int_0^\infty e^{-nx}\frac{\cos(xt)}{x}\,dx\,dt$$

Now I'm stuck because I've never seen anyone performing these,

• $\mathcal{L}\left\{\frac{\cos(at)}{t}\right\}$, although I get this but it scares me that I will need to proceed with a series involving a logarithm in it

• $\mathcal{L}\left\{\frac{\sin(at)}{t^2}\right\}$

So how do I deal with this?

Note : I'm not looking for alternate methods to solve the original integral.

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Since the transform exists for the combination, here's my work so far,

$$2I=\int_0^\infty e^{-2\pi t} \int_0^\infty\frac{1}{e^x-1}\frac{\sin(xt)}{x^2}\,dx\,dt-\int_0^\infty te^{-2\pi t} \int_0^\infty\frac{1}{e^x-1}\frac{\cos(xt)}{x}\,dx\,dt$$

$$\color{red}{\sum_{n=1}^\infty e^{-nx}=\frac{1}{e^x-1}}$$

$$2I=\int_0^\infty e^{-2\pi t} \sum_{n=1}^\infty\int_0^\infty e^{-nx}\frac{\sin(xt)}{x^2}\,dx\,dt-\int_0^\infty te^{-2\pi t} \sum_{n=1}^\infty\int_0^\infty e^{-nx}\frac{\cos(xt)}{x}\,dx\,dt$$

$$2I=\int_0^\infty e^{-2\pi t} \sum_{n=1}^\infty\int_0^\infty e^{-nx}\left(\frac{\sin(xt)}{x^2}-\frac{t\cos(xt)}{x}\right)\,dx\,dt$$

$$I=\int_0^\infty t e^{-2\pi t} \sum_{n=1}^\infty \frac{1}{(n^2+t^2)^2}\,dt$$

Using this, we can differentiate the below sum,

$$\color{red}{\sum_{n=1}^\infty \frac{1}{n^2+t^2} =\frac12\left(\frac{\pi t \coth(\pi t)}{t^2}-\frac{1}{t^2}\right)}$$

$$\color{red}{\sum_{n=1}^{\infty} \frac{1}{(n^2+t^2)^2} = \frac{t^2\,\pi^2\,{\rm csch}^2(t\pi) + t\pi\,{\rm coth}(t\pi) -2}{4t^4}=\frac{\pi^2}{4}\frac{{\rm csch}^2(t\pi)}{t^2}+\frac{\pi}{4}\frac{{\rm coth}(t\pi)}{t^3}-\frac12 \frac{1}{t^4}}$$

$$I=\int_0^\infty t e^{-2\pi t} \left(\frac{\pi^2}{4}\frac{{\rm csch}^2(t\pi)}{t^2}+\frac{\pi}{4}\frac{{\rm coth}(t\pi)}{t^3}-\frac12 \frac{1}{t^4}\right)\,dt$$

$$I=\frac{\pi^2}4\int_0^\infty t^{-1} e^{-2\pi t}\rm csch^2(t\pi)\,dt+\frac{\pi}4\int_0^\infty t^{-2} e^{-2\pi t}\coth(t\pi)\,dt-\frac12\int_0^\infty t^{-3} e^{-2\pi t}\,dt$$

$$I=\int_0^\infty\frac{\pi^2}4 t^{-1} e^{-2\pi t}\rm csch^2(t\pi)\,dt+\frac{\pi}4 t^{-2} e^{-2\pi t}\coth(t\pi)-\frac12t^{-3} e^{-2\pi t}\,dt$$

Leads to another three difficult integrals (with the third one being divergent-?!)

Answer 1

I have found the Laplace transform of $\frac{\cos t}{t}$ in the past. Note that it does not exist unless you either treat it as a generalized function (aka distribution), or equivalently you take the finite part of the Laplace transform integral. consider

$$F(s) = \lim_{\epsilon\rightarrow 0^+} \int_\epsilon^\infty e^{-st} \frac{\cos at}{t} dt $$

Integrating by parts (to turn $1/t$ into $\ln t$) yields integrals that converge but an extra additive term that blows up as $\epsilon\rightarrow 0^+$ The finite part is then obtained by simply throwing out the divergent term. After that I used Taylor series for sinusoids to integrate term by term, and then properties of Gamma and digamma functions to get the final form. In the end I got the same answer as Wolfram Alpha.

I suspect the other example will need similar kinds of tricks.

EDIT: from my comment below, once you know the first Laplace transform then the second should follow after you compute the Laplace transform of $\frac{\sin at}{t}$.

If you are comfortable with the idea of Pseudofunctions from distribution theory then the derivation is very straightforward. Instead of taking the finite part of the Laplace transform integral, we consider $\frac{\cos at}{t} \, u(t)$ to be a pseudofunction denoted by $\mathrm{Pf} \, \frac{\cos at}{t} \, u(t)$; here $u(t) = 1$ for $t>0$ and is zero otherwise. For distributions the standard properties of generalized functions apply.

It is straightforward to show $$ \begin{eqnarray*} \mathcal{L}\left[u(t)\, \ln t\right] & = & -\frac{\gamma + \ln s}{s} \end{eqnarray*} $$ where we have used \begin{eqnarray*} \gamma & = & -\int_0^\infty e^{-t} \, \ln t \end{eqnarray*} We then use the fact that $\mathrm{Pf}\, \frac{u(t)}{t} \equiv \frac{d}{dt} u(t) \, \ln(t)$ and therefore define \begin{eqnarray*} G(s) & \equiv & \mathcal{L}\left[ \mathrm{Pf} \, \frac{u(t)}{t}\right] \\ & = & \mathcal{L}\left[\frac{d}{dt} u(t) \, \ln(t)\right] \\ & = & s \, \mathcal{L}\left[u(t) \, \ln(t)\right] \\ & = & -\gamma - \ln s \end{eqnarray*} Now we note that $$ \begin{eqnarray*} F(s) & = & \mathcal{L}\left[\mathrm{Pf} \, \frac{u(t) \, \cos at}{t}\right] \\ & = & \mathcal{L}\left[\frac{1}{2} \mathrm{Pf}\, \frac{u(t)\, e^{iat}}{t}\right] + \mathcal{L}\left[\frac{1}{2} \mathrm{Pf}\, \frac{u(t)\, e^{-iat}}{t}\right] \\ & = & \frac{1}{2}G(s-ia) + \frac{1}{2} G(s+ia) \\ & = & -\gamma - \frac{1}{2}\ln\left(s^2 + a^2\right) \end{eqnarray*} $$

This should converge in a right-half of the complex $s$ plane. For complex $a$ the region of convergence is $Re(s)>|Im(a)|$.