Evaluation Of Integrals

[Tsakanikas Nickos]

Evaluate the following integrals:

\( \text{(i)} \displaystyle \int_{0}^\infty \int_{0}^{t} \, \frac{ {e}^{-t}\sin{u} }{u} \mathrm{d}u \mathrm{d}t \)

\( \text{(ii)} \displaystyle \int_{0}^\infty \, \frac{ \sin^2{t} }{ t^2 } \mathrm{d}t \)

[Grigorios Kostakos]

\(\rm(i)\) We have that \begin{align*}
\sin{u}&=\displaystyle\mathop{\sum}\limits_{n=0}^{\infty}{\frac{(-1)^nu^{2n+1}}{(2n+1)!}}\quad\Rightarrow\\
\frac{\sin{u}}{u}&=\displaystyle\mathop{\sum}\limits_{n=0}^{\infty}{\frac{(-1)^nu^{2n}}{(2n+1)!}}\quad\Rightarrow\\
\displaystyle\int_{0}^{t}{\frac{\sin{u}}{u}\,du}&=\int_{0}^{t}{\mathop{\sum}\limits_{n=0}^{\infty}{\frac{(-1)^nu^{2n}}{(2n+1)!}}\,du}\\
&=\mathop{\sum}\limits_{n=0}^{\infty}\biggl({\frac{(-1)^n}{(2n+1)!}\int_{0}^{t}u^{2n}\,du}\biggr)\\
&=\mathop{\sum}\limits_{n=0}^{\infty}{\frac{(-1)^n}{(2n+1)!}\,\frac{t^{2n+1}}{2n+1}}\quad\Rightarrow\\
\displaystyle\int_{0}^{t}{\frac{\sin{u}}{u}\,du}&=\mathop{\sum}\limits_{n=0}^{\infty}{\frac{(-1)^nt^{2n+1}}{(2n+1)!(2n+1)}}\quad(1)
\end{align*} So \begin{align*}
\displaystyle \int_{0}^{\infty}{\int_{0}^{t}\frac{{e}^{-t}\sin{u}}{u}\,du \,dt}&=\int_{0}^{\infty}{{e}^{-t}\int_{0}^{t}{\frac{\sin{u}}{u}\,du}\,dt}\\
&\stackrel{(1)}{=}\int_{0}^{\infty}{{e}^{-t}\mathop{\sum}\limits_{n=0}^{\infty}{\frac{(-1)^nt^{2n+1}}{(2n+1)!(2n+1)}}\,dt}\\
&=\mathop{\sum}\limits_{n=0}^{\infty}\biggl({\frac{(-1)^n}{(2n+1)!(2n+1)}\int_{0}^{\infty}{{e}^{-t}t^{2n+1}\,dt}}\biggr)\\
&=\mathop{\sum}\limits_{n=0}^{\infty}{\frac{(-1)^n}{(2n+1)!(2n+1)}\,(2n+1)!}\\
&=\mathop{\sum}\limits_{n=0}^{\infty}{\frac{(-1)^n}{2n+1}}\\
&\stackrel{(*)}{=}\frac{\pi}{4}\,.
\end{align*}

\(\begin{aligned}
(*)\quad&\arctan{z}=\displaystyle\mathop{\sum}\limits_{n=0}^{\infty}{\frac{(-1)^{n}z^n}{2n+1}}\,,\quad z\in{\mathbb{C}}\,,\;|z|\leqslant1\quad\Rightarrow\\
&\frac{\pi}{4}=\arctan{1}=\displaystyle\mathop{\sum}\limits_{n=0}^{\infty}{\frac{(-1)^{n}}{2n+1}}\,.
\end{aligned}\)

\(\rm(ii)\) We will use [easy to prove] that \[\displaystyle \int_{0}^{\infty}{e^{-xt}\sin({2t})\,dt}=\frac{2}{x^2+4}\quad(1)\] So
\begin{align*}
\displaystyle \int_{0}^{\infty}{\frac{\sin^2{t}}{t^2}\,dt}&=\int_{0}^{\infty}{\Bigl({-\frac{1}{t}}\Bigr)'\sin^2{t}\,dt}\\
&=\Bigl[{-\frac{\sin^2{t}}{t}}\Bigr]_{0}^{\infty}+\int_{0}^{\infty}{\frac{2\sin{t}\cos{t}}{t}\,dt}\\
&=\int_{0}^{\infty}{\frac{\sin({2t})}{t}\,dt}\\
&=\int_{0}^{\infty}{\biggl(\int_{0}^{\infty}{e^{-xt}\,dx}\biggr)\sin({2t})\,dt}\\
&=\int_{0}^{\infty}{\biggl(\int_{0}^{\infty}{e^{-xt}\sin({2t})\,dx}\biggr)\,dt}\\
&=\int_{0}^{\infty}{\biggl(\int_{0}^{\infty}{e^{-xt}\sin({2t})\,dt}\biggr)\,dx}\\
&\stackrel{(1)}{=}\int_{0}^{\infty}{\frac{2}{x^2+4}\,dx}\\
&=\frac{\pi}{2}\,.
\end{align*}

[Tsakanikas Nickos]

Here is another solution of (i):
By changing the order of integration, we have that
\begin{align*}
\displaystyle
I = \int_{0}^{\infty}\int_{0}^{t} \frac{{e}^{-t}\sin{u}}{u} \mathrm{d}u\mathrm{d}t &=
\int_{0}^{\infty}\int_{u}^{\infty} \frac{{e}^{-t}\sin{u}}{u} \mathrm{d}t\mathrm{d}u \\
&= \int_{0}^{\infty} \frac{\sin{u}}{u} \left[ -{e}^{-t} |_{u}^{\infty} \right] \mathrm{d}u \\
&= \int_{0}^{\infty} {e}^{-1u} \frac{\sin{u}}{u} \mathrm{d}u \\
&= F(1),
\end{align*}
where \( \displaystyle F(s) = \mathcal{L} \left\{ \frac{\sin{t}}{t} \right\}. \)
Since \( \displaystyle \lim_{t \to 0} \frac{\sin{t}}{t} = 1 \in \mathbb{R}, \text{ we have that } \)
\begin{align*}
F(s) = \mathcal{L} \left\{ \frac{\sin{t}}{t} \right\} &= \int_{s}^{\infty} \mathcal{L} \left\{\sin{t}\right\} \mathrm{d}u \\
&= \int_{s}^{\infty} \frac{1}{u^2+1} \mathrm{d}u \\
&= \arctan{u}|_{s}^{\infty} \\
&= \frac{\pi}{2} - \arctan{s}.
\end{align*}
Thus, \( \displaystyle I = F(1) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \)