Evaluate $\int_{0}^{\infty}\frac{\sin^{2}\left ( t \right )}{t^{2}}dt$ with help of Laplace transform

Question

Using the following identity $$\int_{0}^{\infty}\frac{f\left ( t \right )}{t}dt= \int_{0}^{\infty}\mathcal{L}\left \{ f\left ( t \right ) \right \}\left ( u \right )du$$ I rewrote $$\int_{0}^{\infty}\frac{\sin^{2}\left ( t \right )}{t^{2}}dt$$ as $$\int_{0}^{\infty}\frac{\sin^{2}\left ( t \right )}{t\cdot t}dt$$ And thus the initial integral should be easily evaluated as $$\int_{0}^{\infty}\mathcal{L}\left \{ \frac{\sin^{2}t}{t} \right \}\left ( u \right )du$$ According to my calculations, this is equal to $$\int_{0}^{\infty}\left ( \frac{1}{4}\ln \left ( u^{2}+4 \right )-\frac{u^{2}}{4} \right )du$$ Which evaluates to $\frac{\pi}{16}$. Being that this acutally a well-known integral and that its value is actually $\frac{\pi}{2}$ I think that I made a crucial mistake somewhere. Any ideas?

Answer 1

You seemed to have confused logarithmic identities. The integrand should be

$$\frac{1}{4} \log \left(\frac{4}{s^2}+1\right)$$

which indeed evaluates as $\pi/2$. I am guessing you wrongly factored out an $s^2$

Answer 2

We have

$$\begin{align} \int_0^{\infty}\frac{\sin^2t}{t}e^{-ut}&=\frac{1}{2}\int_0^{\infty}\frac{1-\cos t}{t}e^{-(u/2)t}dt\\\\ &=\frac12\log\left(\frac{(u/2)^2+1}{(u/2)^2}\right)\\\\ &=\frac12 \log\left(\frac{u^2+4}{u^2}\right) \end{align}$$