I am trying to show $\int^\infty_0\frac{\sin(x)}{x}dx=\frac{\pi}{2}$
Solution 1:
Since the integrand is even function, then you can consider
$$ \int^\infty_0\frac{\sin(x)}{x}dx = \frac{1}{2}\int_{-\infty}^\infty \frac{\sin(x)}{x}dx .$$
Solution 2:
Though this is not what you asked for originally, let me offer a a very elegant real variable approach, for a change. I actually wrote an extensive blog post on this, with a few nice graphs, a while ago, so here I'll just give you the bare bones sketch. The argument comes from Introduction to Calculus and Analysis, vol. I, by Richard Courant and Fritz John, Interscience Publishers (1965), reprinted by Springer (1989).
Note first that $\displaystyle\int_0^\infty\frac{\sin x}x\,dx=\sum_{k=0}^\infty a_k$, where $\displaystyle a_k=\int_{\pi k}^{\pi(k+1)}\frac{\sin x}x\,dx=\int_0^\pi\frac{(-1)^k\sin t}{\pi k +t}\,dt$. This shows that the integral converges, by Leibniz criterion.
Second, note that $\displaystyle\int_0^\infty\frac{\sin x}x\,dx=\lim_{\rho\to\infty}\int_0^{\pi\rho}\frac{\sin x}{x}\,dx=\lim_{\rho\to\infty}\int_0^\pi\frac{\sin(\rho t)}{t}\,dt$.
Third, and this is the key, $\displaystyle \lim_{k\to\infty}\int_0^\pi\sin\left(\left(k+\frac12\right)t\right)\left(\frac1t-\frac1{2\sin(t/2)}\right)\,dt=0$. To see this, let $\displaystyle f(t)=\frac1t-\frac1{2\sin(t/2)}$. Check that $\lim_{t\to0^+}f(t)=0$ and, defining $f(0)=0$, we have that $f'(0)$ exists and equals $-1/24=\lim_{t\to0^+}f'(t)$. It follows that $f$ is continuously differentiable on $[0,\pi]$ so we can integrate by parts to get $$ \int_0^\pi\sin((k+1/2)t)f(t)\,dt=\frac1{k+1/2}\int_0^\pi\cos((k+1/2)t)f'(t)\,dt\to_{k\to\infty}0. $$
It follows that $\displaystyle\int_0^\infty\frac{\sin x}x\,dx=\lim_{k\to\infty}\int_0^\pi\frac{\sin(k+\frac12)x}{2\sin(x/2)}\,dx$.
Fourth, recall the identity $\displaystyle f_k(t)=\frac{\sin(k+\frac12)t}{2\sin(t/2)}$, where $$ f_k(t)=\frac12+\cos t+\cos(2t)+\dots+\cos(kt). $$ It follows that $\displaystyle \int_0^\pi\frac{\sin(k+\frac12)t}{2\sin(t/2)}\,dt=\int_0^\pi f_k(t)\,dt= \frac{\pi}2$, and we are done.
Solution 3:
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
- \begin{align}\color{#0000ff}{\large% \int_{-\infty}^{\infty}{\sin\pars{x} \over x}\,\dd x} &= \int_{-\infty}^{\infty}\pars{\half\int_{-1}^{1}\expo{\ic k x}\,\dd k}\,\dd x \\[5mm] & = \pi\int_{-1}^{1}\pars{\int_{-\infty}^{\infty}\expo{\ic k x}\,{\dd x \over 2\pi}} \,\dd k = \pi\int_{-1}^{1}\delta\pars{k} = \color{#0000ff}{\Large\pi} \end{align}
- \begin{align} 0& =\int_{-R}^{-\epsilon}{\expo{\ic x} \over x}\,\dd x + \int_{\pi}^{0}{\expo{\ic\epsilon\expo{\ic\theta}} \over \epsilon\expo{\ic\theta}}\, \epsilon\expo{\ic\theta}\ic\,\dd\theta + \int_{\epsilon}^{R}{\expo{\ic x} \over x}\,\dd x + \int_{0}^{\pi}{\expo{\ic R\expo{\ic\theta}} \over R\expo{\ic\theta}}\, R\expo{\ic\theta}\ic\,\dd\theta \end{align} With the limit $\epsilon \to 0^{+}$: \begin{align} 0& =\lim_{\epsilon \to 0^{+}}\pars{\int_{-R}^{-\epsilon}{\expo{\ic x} \over x}\,\dd x + \int_{-\epsilon}^{-R}{\expo{-\ic x} \over x}\,\dd x} - \ic\pi \\[2mm] & + \ic\int_{0}^{\pi}\expo{\ic R\expo{\ic\theta}}\,\dd\theta \\[3mm]&= -2\ic\int_{0}^{R}{\sin\pars{x} \over x}\,\dd x -\ic\pi + \ic\int_{0}^{\pi}\expo{\ic R\expo{\ic\theta}}\,\dd\theta \\[5mm] \color{#0000ff}{\large% \int_{0}^{\infty}{\sin\pars{x} \over x}\,\dd x} &= \color{#0000ff}{\large{\pi \over 2}} + {1 \over 2}\overbrace{\lim_{R \to \infty} \int_{0}^{\pi}\expo{\ic R\cos\pars{\theta}}\expo{-R\sin\pars{\theta}}\,\dd\theta} ^{\ds{=\ 0}} \end{align}