Solving the integral without using Laplace

Question

On my last question, i was trying to find the Laplace of the following function, $$\int_0^\infty \frac{e^{-2t}\sinh t\sin t }{t} dt.$$
Now I am wondering, can I solve the integral directly without using the Laplace, could someone provide me some hints to solve the integral, I am having some trouble where to start from.
Well, I am thinking of writing the $\sinh{t}$ and $\sin{t}$ in their exponential form or is there any other way to solve this question?

Answer 1

Consider the following integral:

\begin{equation} I \,=\displaystyle{\int\limits_{0}^{+\infty}} \frac{e^{-2t}\sinh(t)\sin(t)}{t}\,dt \end{equation}

We can simplify the integrand if we transform $\sinh(t)$ into its exponential form, given that:

\begin{equation} \sinh(t)= \frac{1-e^{-2t}}{2e^{-t}} \end{equation}

So, our integrand now has the following form

\begin{equation} I \,=\displaystyle{\int\limits_{0}^{+\infty}} \frac{e^{-2t}\sin(t)(1-e^{-2t})}{2te^{-t}}\,dt \,\,=\displaystyle{\int\limits_{0}^{+\infty}} \frac{\sin(t) - e^{-2t}\sin(t)}{2te^{t}}\,dt \end{equation}

Now, separating the integrand leaves us with two very similar integrals to compute:

\begin{equation} I\,=\displaystyle{\int\limits_{0}^{+\infty}} \frac{\sin(t) - e^{-2t}\sin(t)}{2te^{t}}\,dt \,\,= \displaystyle{\int\limits_{0}^{+\infty}} \frac{e^{-t}\sin(t)}{2t}\,dt \,\,- \displaystyle{\int\limits_{0}^{+\infty}} \frac{e^{-3t}\sin(t)}{2t}\,dt \end{equation}

Let's consider the following integral:

\begin{equation} I(\alpha)\,=\displaystyle{\int\limits_{0}^{+\infty}} \frac{e^{-\alpha t}\sin(t)}{2t}\,dt \,\,\,\,\,\,\forall \alpha>0 \end{equation}

It is clear that we can express our original integral $I$ in terms of $I(\alpha)$, such that:

\begin{equation} I = I(\alpha = 1) - I(\alpha = 3) \end{equation}

We can compute $I(\alpha)$ using the Leibniz rule, or what it is sometimes called differentiation under the integral sign. Now, let's differentiate $I(\alpha)$ with respect to $\alpha$:

\begin{equation} \frac{dI(\alpha)}{d\alpha} = \frac{d}{d\alpha}\left[\displaystyle{\int\limits_{0}^{+\infty}} \frac{e^{-\alpha t}\sin(t)}{2t}\,dt\right] \end{equation}

Given that the upper and lower bounds are not dependent in our parameter $\alpha$, by the Leibniz rule, we can introduce the ordinary derivative into the integrand, transforming it into a parcial derivative with respect to our new parameter:

\begin{equation} I'(\alpha) = \displaystyle{\int\limits_{0}^{+\infty}} \frac{\partial}{\partial\alpha}\left[\frac{e^{-\alpha t}\sin(t)}{2t}\right]\,dt = \displaystyle{\int\limits_{0}^{+\infty}} \frac{e^{-\alpha t}\sin(t)(-t)}{2t}\,\, dt = -\frac{1}{2}\displaystyle{\int\limits_{0}^{+\infty}} e^{-\alpha t}\sin(t)\,\, dt \end{equation}

The last integral is fairly easy to integrate, as it can be solved integrating by parts twice. This cyclical integral yields the following expression:

\begin{equation} \displaystyle{\int\limits_{0}^{+\infty}} e^{-\alpha t}\sin(t)\,\, dt = \frac{1}{1 + \alpha^{2}} \end{equation}

So, we now have a new expression for $I(\alpha)$:

\begin{equation} I'(\alpha) = -\frac{1}{2\left(1 + \alpha^{2}\right)} \end{equation}

Now, we would like to integrate $I'(\alpha)$ with respect to $\alpha$, in order to go back to our original expression. To do so, we need to find appropiate integration limits. We would like to find a limit in which our integrand goes to zero, and another one that allows us to go back to our initial integral. Notice that the integrand vanishes when $\alpha \rightarrow +\infty$, given that the definite integral of zero over any inteval, must be zero, then $I(\alpha \rightarrow +\infty)=0$, so we will let our upper bound approach infinity. We can let our lower bound be $\alpha$.

Let's integrate both sides of the equation with respect to $\alpha$, in the interval $(\alpha, \infty)$:

\begin{equation} \displaystyle{\int\limits_{\alpha}^{+\infty}} I'(\alpha) = \underbrace{I(\alpha \rightarrow +\infty)}_{zero} - I(\alpha) = -\frac{1}{2}\displaystyle{\int\limits_{\alpha}^{+\infty}} \frac{d\alpha}{1 + \alpha^{2}} \end{equation}

\begin{equation} I(\alpha) = \frac{1}{2}\displaystyle{\int\limits_{\alpha}^{+\infty}} \frac{d\alpha}{1 + \alpha^{2}} = \frac{1}{2}\Big[\arctan(\alpha \rightarrow +\infty) - \arctan(\alpha)\Big] \end{equation}

Finally, we obtain an expression for $I(\alpha)$:

\begin{equation} I(\alpha) = \frac{1}{2}\bigg[\frac{\pi}{2} - \arctan(\alpha)\bigg] \end{equation}

We can further simplify the last expression if we consider the following identity:

\begin{equation} \arctan\left(\frac{1}{\alpha}\right) = \frac{\pi}{2} - \arctan(\alpha) = \mathrm{arccot}(\alpha) \,\,\,\,\, \forall \alpha>0 \end{equation}

So, our initial integral $I(\alpha)$ can be expressed as follows:

\begin{equation} I(\alpha)\,=\displaystyle{\int\limits_{0}^{+\infty}} \frac{e^{-\alpha t}\sin(t)}{2t}\,dt = \frac{1}{2}\mathrm{arccot(\alpha)} \end{equation}

With this result, we can finally compute our original integral $I$:

\begin{equation} I = I(\alpha = 1) - I(\alpha = 3) = \frac{1}{2}\bigg[\mathrm{arccot(1)} - \mathrm{arccot(3)}\bigg] \end{equation}

\begin{equation} I \,=\displaystyle{\int\limits_{0}^{+\infty}} \frac{e^{-2t}\sinh(t)\sin(t)}{t}\,dt = \frac{\pi}{8} - \frac{1}{2}\mathrm{arccot(3)} \approx 0,231823\,... \end{equation}

The result given by Wolfram-Alpha

Answer 2

Apply Frullani integral theorem twice (see Wikipedia). The OP's integral can be written as $$I=\frac{1}{2}\Big( \frac{1}{2i}\int_0^\infty \big( e^{-t(1-i)} - e^{-t(1+i)}\big) \frac{dt}{t} -\frac{1}{2i}\int_0^\infty \big( e^{-t(3-i)} - e^{-t(3+i)}\big) \frac{dt}{t}\Big)$$ $$=\frac{1}{2}\Big( \frac{1}{2i} \log{\big((1+i)/(1-i)\big)} - \frac{1}{2i} \log{\big((3+i)/(3-i)\big)} \Big)$$ $$= \frac{1}{2}\big( \frac{\pi}{4} - \frac{1}{2} \tan^{-1}(3/4) \big) $$