I am calculating an integral $\displaystyle \int \limits_{-\infty}^{\infty}\dfrac{\sin(x)}{x}$ and I dont seem to be getting an answer.
When I integrate by parts twice, I get:
$$\displaystyle \int \limits _{-\infty}^{\infty}\frac{\sin(x)}{x}dx = \left[\frac{\sin(x)\ln(x) - \frac{\cos(x)}{x}}{2}\right ]_{-\infty}^{+\infty}$$
What will be the answer to that ?
Answer
Hint: From the viewpoint of improper Lebesgue integrals or in sense of Cauchy principal value is integral is legitimate. Integration by parts.
\begin{align}
\int \limits_{-\infty}^{\infty}\dfrac{\sin(x)}{x} \mbox{d} x
=
&
\lim_{t\to\infty}\int \limits_{-t}^{\frac{1}{t}}\dfrac{\sin(x)}{x} \mbox{d} x
+
\lim_{t\to\infty}\int \limits_{\frac{1}{t}}^{t}\dfrac{\sin(x)}{x} \mbox{d} x
\\
=
&
\lim_{t\to\infty}\int \limits_{-t}^{\frac{1}{t}}\sin(x)(\log x)^\prime \mbox{d} x
+
\lim_{t\to\infty}\int \limits_{\frac{1}{t}}^{t}\sin(x)(\log x)^\prime\mbox{d} x
\\
\end{align}
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