I am trying to solve the following integrals:
1.) $$\int_0^{\infty} \frac{\sin (t)(\sin (t)-t)}{t^4}dt$$
and
2.) $$\int_0^{\infty} \frac{\sin (t)\left(\sin ^2(t)-t^2\right)}{t^5} d t$$
I tried multiplying throughout and splitting the integral so that we could use the integral of the $\operatorname{sinc}$ function and its powers, but they don't converge when we do so.
I then used WolframAlpha (1, 2), which gave the answers to be $-\frac{\pi}{12}$ and $-\frac{5 \pi}{32}$ respectively.
How does one go about solving these integrals? Any help will be appreciated!
Thank You!
EDIT: In this post, the OP has tried generalizing the problem for different powers of
the sine function thus-
$$I(a,b,c,d)=\int_0^{\infty} \frac{\sin^a (t)\left(\sin ^b(t)-t^c\right)}{t^d} d t.$$
It should be noted that the integral converges for some values of $a,b,c$ and $d.$ I am still trying to figure out when the integral converges.
For some other cases where I found the integral to converge, I tried using the methods already given here(CHAMSI's and Random Variable's) and was able to evaluate a few, but it gets more complicated as we go for higher powers. In particular, I was able to evaluate $I(3,1,1,6)=-\frac{29 \pi}{480}$ and $I(2,2,2,6)=-\frac{2 \pi}{15}.$
My question now is to explicitly evaluate $I(a,b,c,d).$
Hope the OP also sees this post.
Thank you!
Best Answer
The first integral we can present in the form $$I_1=\frac12\Im\int_{-\infty}^\infty\frac{e^{it}(\sin t-t)}{t^4}dt$$ where we evaluate integral in the principal value sense. We close the contour by a big arch in the upper half of the complex plane (going counter-clockwise).
Taking $z=x+iy$, in the upper half-plane $\, |e^{iz}\sin z|=\Big|e^{-y+ix}\frac{e^{-y+ix}-e^{y-ix}}{2i}\Big|=\frac12\big|1-e^{-2y+2ix}\big|<1$. It means the integral along a big arch $\to0$ as $R\to\infty$. Adding also the small arch of the radius $r\to 0$ above the point $z=0$ (clockwise, to close the contour), we have $$\frac12\Im\oint\frac{e^{iz}(\sin z-z)}{z^4}dz=I_1+\frac12\Im\int_{C_r}\frac{e^{iz}(\sin z-z)}{z^4}dz=0$$ because we do not have poles inside the contour. Hence, $$I_1=-\frac12\Im\int_{C_r}\frac{e^{iz}(\sin z-z)}{z^4}dz=\frac12\Im\,\left(\pi i\underset{z=0}{\operatorname{Res}}\frac{e^{iz}(\sin z-z)}{z^4}\right)=-\frac\pi{12}$$ We cannot evaluate the second integral in the same way, because $e^{iz}\sin^2z$ is growing exponentially in the upper and lower half-planes (though, we could use this approche to evaluate, for example, $\,\int_0^{\infty} \frac{\sin (at)\left(\sin ^2t-t^2\right)}{t^5} dt,\,\,\text{where}\,\,a\geqslant2$).
Making some transformations of the integrand $$I_2=\frac12\Im\int_{-\infty}^\infty \frac{e^{it}\left(\sin ^2t-t^2\right)}{t^5} dt=\frac14\Im\int_{-\infty}^\infty \frac{e^{it}\left(1-\cos(2t)-2t^2\right)}{t^5} dt$$ $$=\frac14\Im\int_{-\infty}^\infty \left(e^{it}-\frac12e^{3it}-\frac12e^{-it}-2t^2e^{it}\right)\frac{dt}{t^5}$$ where we evaluate the integral in the principal value sense. It is straightforward to present the integral in the following form: $$I_2=\frac14\Im\int_{-\infty}^\infty \Big(e^{it}-\frac12e^{3it}-2t^2e^{it}+\frac{it}2-\frac{it^3}{12}\Big)\frac{dt}{t^5}$$ $$-\frac14\Im\int_{-\infty}^\infty \Big(\frac12e^{-it}+\frac{it}2-\frac{it^3}{12}\Big)\frac{dt}{t^5}$$ We see that every integrand has a simple pole at $t=0$. Now we use the same approach as for the integral $I_1$, closing the contour by means of small and big arches. For the first integral we close the contour in the upper half-plane. For the second one we close the contour in the lower half-plane, adding the small arch in this case below $z=0$, counter-clockwise, (what gives an additional minus). Decomposing the integrands near $t=0$ and gathering $\,t^4\,$ powers, finally we get $$I_2=\frac14\Im\,\frac{\pi i}{2\cdot4!}\Big(\big(2-3^4+48\big)+1\Big)=-\frac{30\pi}{8\cdot6\cdot4}=-\frac{5\pi}{32}$$