my first post and a cool and challenging integral
my first post and a cool and challenging integral
Evaluate:
$$\int_{0}^{\infty}\frac{(1-\sin(ax))(1-\cos(bx))}{x^{2}}dx$$
$$\int_{0}^{\infty}\frac{(1-\sin(ax))(1-\cos(bx))}{x^{2}}dx$$
Re: my first post and a cool and challenging integral
Maybe I am missing something simple or direct
\begin{align*}I&=\int_0^{\infty} \frac{(1-\sin ax)(1-\cos bx)}{x^2}\,dx\\&= \int_0^{\infty} \frac{1-\cos bx - \sin ax + \frac{1}{2}\sin (a+b)x + \frac{1}{2}\sin (a-b)x}{x^2}\,dx\\&= 2\int_0^{\infty} \frac{\sin^2 \frac{bx}{2}}{x^2}\,dx - \int_0^{\infty}\left(\sin ax - \frac{1}{2}\sin (a+b)x - \frac{1}{2}\sin (a-b)x\right)\frac{\,dx}{x^2}\\&= \frac{\pi b}{2} + \left.\frac{\left(\sin ax - \frac{1}{2}\sin (a+b)x - \frac{1}{2}\sin (a-b)x\right)}{x}\right\vert_0^{\infty} - \int_0^{\infty} \left(a\cos ax -\frac{a+b}{2}\cos (a+b)x - \frac{a-b}{2}\cos (a-b)x\right)\frac{\,dx}{x}\\&= \frac{\pi b}{2} - \int_0^{\infty} \left(a\cos ax -\frac{a+b}{2}\cos (a+b)x - \frac{a-b}{2}\cos (a-b)x\right)\frac{\,dx}{x} \end{align*}
To compute the above integral we recall that: $\displaystyle \int_0^{\infty} x^{s-1}e^{ikx} \,dx= k^{-s}e^{i\pi s/2}\Gamma (s)$ for $k > 0$.
\begin{align*}&\int_0^{\infty} \left(a\cos ax -\frac{a+b}{2}\cos (a+b)x - \frac{a-b}{2}\cos (a-b)x\right)x^{s-1}\,dx \\&= \left(a^{1-s}-\frac{1}{2}(a+b)^{1-s} - \frac{1}{2}(a-b)^{1-s}\right)\cos \left(\frac{\pi s}{2}\right)\Gamma (s)\end{align*}
Thus, \begin{align*}&\int_0^{\infty} \left(a\cos ax -\frac{a+b}{2}\cos (a+b)x - \frac{a-b}{2}\cos (a-b)x\right)\frac{\,dx}{x}\\&= \lim\limits_{s \to 0^{+}}\frac{1}{s} \left(a^{1-s}-\frac{1}{2}(a+b)^{1-s} - \frac{1}{2}(a-b)^{1-s}\right)\\&= -a\log a + \frac{1}{2}(a+b)\log (a+b)+\frac{1}{2}(a-b)\log (a-b)\end{align*}
Hence, $$I = \frac{\pi b}{2} + a\log a - \frac{1}{2}(a+b)\log (a+b)-\frac{1}{2}(a-b)\log (a-b)$$
\begin{align*}I&=\int_0^{\infty} \frac{(1-\sin ax)(1-\cos bx)}{x^2}\,dx\\&= \int_0^{\infty} \frac{1-\cos bx - \sin ax + \frac{1}{2}\sin (a+b)x + \frac{1}{2}\sin (a-b)x}{x^2}\,dx\\&= 2\int_0^{\infty} \frac{\sin^2 \frac{bx}{2}}{x^2}\,dx - \int_0^{\infty}\left(\sin ax - \frac{1}{2}\sin (a+b)x - \frac{1}{2}\sin (a-b)x\right)\frac{\,dx}{x^2}\\&= \frac{\pi b}{2} + \left.\frac{\left(\sin ax - \frac{1}{2}\sin (a+b)x - \frac{1}{2}\sin (a-b)x\right)}{x}\right\vert_0^{\infty} - \int_0^{\infty} \left(a\cos ax -\frac{a+b}{2}\cos (a+b)x - \frac{a-b}{2}\cos (a-b)x\right)\frac{\,dx}{x}\\&= \frac{\pi b}{2} - \int_0^{\infty} \left(a\cos ax -\frac{a+b}{2}\cos (a+b)x - \frac{a-b}{2}\cos (a-b)x\right)\frac{\,dx}{x} \end{align*}
To compute the above integral we recall that: $\displaystyle \int_0^{\infty} x^{s-1}e^{ikx} \,dx= k^{-s}e^{i\pi s/2}\Gamma (s)$ for $k > 0$.
\begin{align*}&\int_0^{\infty} \left(a\cos ax -\frac{a+b}{2}\cos (a+b)x - \frac{a-b}{2}\cos (a-b)x\right)x^{s-1}\,dx \\&= \left(a^{1-s}-\frac{1}{2}(a+b)^{1-s} - \frac{1}{2}(a-b)^{1-s}\right)\cos \left(\frac{\pi s}{2}\right)\Gamma (s)\end{align*}
Thus, \begin{align*}&\int_0^{\infty} \left(a\cos ax -\frac{a+b}{2}\cos (a+b)x - \frac{a-b}{2}\cos (a-b)x\right)\frac{\,dx}{x}\\&= \lim\limits_{s \to 0^{+}}\frac{1}{s} \left(a^{1-s}-\frac{1}{2}(a+b)^{1-s} - \frac{1}{2}(a-b)^{1-s}\right)\\&= -a\log a + \frac{1}{2}(a+b)\log (a+b)+\frac{1}{2}(a-b)\log (a-b)\end{align*}
Hence, $$I = \frac{\pi b}{2} + a\log a - \frac{1}{2}(a+b)\log (a+b)-\frac{1}{2}(a-b)\log (a-b)$$
Last edited by r9m on Sun Dec 20, 2015 1:09 am, edited 1 time in total.
Re: my first post and a cool and challenging integral
Clever and consise, RD.
Re: my first post and a cool and challenging integral
This in principle is not too far from my first attempt, but I am adding it for the sake of variety:
\begin{align*}&\int_0^{\infty} \frac{(1-\sin ax)(1-\cos bx)}{x^2}\,dx\\&= \int_0^{\infty} (1-\sin ax)(1-\cos bx) \int_0^{\infty} ye^{-yx}\,dy\,dx\\&= \int_0^{\infty} \int_0^{\infty} ye^{-yx}\left(1-\sin ax -\cos bx + \frac{1}{2}\sin (a+b)x + \frac{1}{2}\sin (a-b) x\right)\,dx\,dy\\&= \int_0^{\infty} y\left(\frac{1}{y}-\frac{a}{a^2+y^2} - \frac{y}{b^2+y^2}+\frac{1}{2}\frac{(a+b)}{(a+b)^2+y^2}+\frac{1}{2}\frac{(a-b)}{(a-b)^2+y^2}\right)\,dy\\&= \frac{\pi b}{2} - \int_0^{\infty} y\left(\frac{a}{a^2+y^2}-\frac{1}{2}\frac{(a+b)}{(a+b)^2+y^2}-\frac{1}{2}\frac{(a-b)}{(a-b)^2+y^2}\right)\,dy\\&= \frac{\pi b}{2} - \frac{1}{2}\int_0^{\infty} \left(\frac{a}{a^2+y}-\frac{1}{2}\frac{(a+b)}{(a+b)^2+y}-\frac{1}{2}\frac{(a-b)}{(a-b)^2+y}\right)\,dy\\&= \frac{\pi b}{2}+a\log a - \frac{1}{2}(a+b)\log (a+b) - \frac{1}{2}(a-b)\log (a-b)\end{align*}
\begin{align*}&\int_0^{\infty} \frac{(1-\sin ax)(1-\cos bx)}{x^2}\,dx\\&= \int_0^{\infty} (1-\sin ax)(1-\cos bx) \int_0^{\infty} ye^{-yx}\,dy\,dx\\&= \int_0^{\infty} \int_0^{\infty} ye^{-yx}\left(1-\sin ax -\cos bx + \frac{1}{2}\sin (a+b)x + \frac{1}{2}\sin (a-b) x\right)\,dx\,dy\\&= \int_0^{\infty} y\left(\frac{1}{y}-\frac{a}{a^2+y^2} - \frac{y}{b^2+y^2}+\frac{1}{2}\frac{(a+b)}{(a+b)^2+y^2}+\frac{1}{2}\frac{(a-b)}{(a-b)^2+y^2}\right)\,dy\\&= \frac{\pi b}{2} - \int_0^{\infty} y\left(\frac{a}{a^2+y^2}-\frac{1}{2}\frac{(a+b)}{(a+b)^2+y^2}-\frac{1}{2}\frac{(a-b)}{(a-b)^2+y^2}\right)\,dy\\&= \frac{\pi b}{2} - \frac{1}{2}\int_0^{\infty} \left(\frac{a}{a^2+y}-\frac{1}{2}\frac{(a+b)}{(a+b)^2+y}-\frac{1}{2}\frac{(a-b)}{(a-b)^2+y}\right)\,dy\\&= \frac{\pi b}{2}+a\log a - \frac{1}{2}(a+b)\log (a+b) - \frac{1}{2}(a-b)\log (a-b)\end{align*}
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