Wednesday, February 20, 2019

integration - A 'complicated' integral: $ int limits_{-infty}^{infty}frac{sin(x)}{x}$




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}


No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...