Tuesday, 11 August 2015

calculus - compute improper integrals using integration by parts



Compute



\begin{equation*}
\int_0^\infty \frac{\sin^4(x)}{x^2}~dx\text{ and }\int_0^\infty \frac{\sin (ax) \cos (bx)}{x}~dx.

\end{equation*}



For the first integral I tried letting $u = \sin ^4 x$ and $dv= \frac{1}{x^2}~dx$, which simplified to $\int_0 ^\infty \frac{4 \sin^3(x) \cos(x)}{x}~dx$ and applying integration by parts again with $u = \sin^3(x) \cos (x)$ and $dv = \frac{1}{x}$, I got



$$\left.\vphantom{\frac11}[\sin ^3 (x) \cos (x) \ln (x)] \right|_0^\infty - \int_0^\infty \ln (x) [3 \sin ^2 (2x) - \sin^4 (x)]~dx$$



The only problem is that it appears that the term $[\sin ^3 (x) \cos (x) \ln (x)] \big\vert_0^\infty$ evaluates to infinity and neither the $3\int_0^\infty \ln (x) \sin ^2 (2x)~dx$ term nor the $3\int_0^\infty \ln (x) \sin^4 (x)~dx$ converges.This looks like a dead end but the other choice of $u$ and $dv$ looks even worse, so I'm not sure how to proceed.



For the second integral I tried $u = \sin (ax) \cos (bx)$ and $dv = \frac{1}{x}$, which gives me $[\sin (ax) \cos (bx) \ln (x)] \big\vert_0^\infty + \int_0^\infty \frac{b-a}{2} \ln (x) \cos ((a-b)x)~dx - \int_0 ^ \infty \frac{b+a}{2} \ln (x) \cos ((a+b)x)~dx$, which also looks like a dead end because all three of the terms go to infinity.




Any help is appreciated.


Answer



For the first one, we have
\begin{align}
\int_0^{\infty} \dfrac{\sin^4(x)}{x^2}dx & = \int_0^{\infty} \dfrac{\sin^2(x)(1-\cos^2(x))}{x^2}dx = \int_0^{\infty} \dfrac{\sin^2(x)}{x^2}dx - \int_0^{\infty} \dfrac{\sin^2(x)\cos^2(x)}{x^2}dx\\
& = \int_0^{\infty} \text{sinc}^2(x) dx - \int_0^{\infty} \dfrac{\sin^2(2x)}{(2x)^2}dx = \int_0^{\infty} \text{sinc}^2(x) dx - \dfrac12\int_0^{\infty} \text{sinc}^2(x) dx\\
& = \dfrac12\int_0^{\infty} \text{sinc}^2(x) dx = \dfrac{\pi}4
\end{align}
From here, we have the value of $\int_0^{\infty} \text{sinc}^m(x)dx$







For the second one, we have
$$\sin(ax)\cos(bx) = \dfrac{\sin((a+b)x) + \sin((a-b)x)}2$$
We hence have
$$\dfrac12\int_0^{\infty}\dfrac{\sin((a+b)x)}xdx + \dfrac12\int_0^{\infty}\dfrac{\sin((a-b)x)}xdx =
\begin{cases}
\dfrac{\pi}2 & \text{ if }a>b\\
\dfrac{\pi}4 & \text{ if }a=b\\
0 & \text{ if }a < b

\end{cases}$$


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