Evaluating two improper integrals involving powers of $\sin(t)$

Question

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!

Answer 1

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 $$I_2=\frac14\Im\,\frac{\pi i}{2\cdot4!}\Big(\big(2-3^4+48\big)+1\Big)=-\frac{30\pi}{8\cdot6\cdot4}=-\frac{5\pi}{32}$$

Answer 2

We will be using the following $\mathrm{sinc}$ integrals : $ \int_{0}^{+\infty}{\frac{\sin^{2}{x}}{x^{2}}\,\mathrm{d}x}=\int_{0}^{+\infty}{\frac{\sin{x}}{x}\,\mathrm{d}x}=\frac{\pi}{2} $, as well as the following formulas for $ x\in\mathbb{R}\setminus\left\lbrace 0\right\rbrace :$

\begin{aligned}\frac{x-\sin{x}}{x^{3}}=\frac{1}{2}\int_{0}^{1}{\left(1-y\right)^{2}\cos{\left(xy\right)}\,\mathrm{d}y}\\ \frac{x^{2}-\sin^{2}{x}}{x^{3}}=2\int_{0}^{1}{\left(1-y\right)^{2}\sin{\left(2xy\right)}\,\mathrm{d}y}\end{aligned}

which can be easily proven using some integration by parts.

We have : \begin{aligned}\int_{0}^{+\infty}{\frac{\sin{x}\left(x-\sin{x}\right)}{x^{4}}\,\mathrm{d}x}&=\frac{1}{2}\int_{0}^{+\infty}{\int_{0}^{1}{\left(1-y\right)^{2}\frac{\sin{x}\cos{\left(xy\right)}}{x}\,\mathrm{d}y}\,\mathrm{d}x}\\ &=\frac{1}{2}\int_{0}^{1}{\left(1-y\right)^{2}\int_{0}^{+\infty}{\frac{\sin{x}\cos{\left(xy\right)}}{x}\,\mathrm{d}x}\,\mathrm{d}y}\\ &=\frac{1}{2}\int_{0}^{1}{\left(1-y\right)^{2}\left(\int_{0}^{+\infty}{\frac{\sin{\left(\left(1-y\right)x\right)}}{2x}\,\mathrm{d}x}+\int_{0}^{+\infty}{\frac{\sin{\left(\left(1+y\right)x\right)}}{2x}\,\mathrm{d}x}\right)\,\mathrm{d}y}\\ &=\frac{1}{2}\int_{0}^{1}{\left(1-y\right)^{2}\left(\frac{1}{2}\int_{0}^{+\infty}{\frac{\sin{u}}{u}\,\mathrm{d}u}+\frac{1}{2}\int_{0}^{+\infty}{\frac{\sin{v}}{v}\,\mathrm{d}v}\right)\,\mathrm{d}y}\\ &=\frac{\pi}{4}\int_{0}^{1}{\left(1-y\right)^{2}\,\mathrm{d}y}\\ &=\frac{\pi}{12}\end{aligned}

Similarly :

\begin{aligned}\int_{0}^{+\infty}{\frac{\sin{x}\left(x^{2}-\sin^{2}{x}\right)}{x^{5}}\,\mathrm{d}x}&=2\int_{0}^{+\infty}{\int_{0}^{1}{\left(1-y\right)^{2}\frac{\sin{x}\sin{\left(2xy\right)}}{x^{2}}\,\mathrm{d}y}\,\mathrm{d}x}\\ &=2\int_{0}^{1}{\left(1-y\right)^{2}\int_{0}^{+\infty}{\frac{\sin{x}\sin{\left(2xy\right)}}{x^{2}}\,\mathrm{d}x}\,\mathrm{d}y}\\ &=2\int_{0}^{1}{\left(1-y\right)^{2}\left(\int_{0}^{+\infty}{\frac{\sin^{2}{\left(\left(y+\frac{1}{2}\right)x\right)}}{x^{2}}\,\mathrm{d}x}-\int_{0}^{+\infty}{\frac{\sin^{2}{\left(\left(y-\frac{1}{2}\right)x\right)}}{x^{2}}\,\mathrm{d}x}\right)\,\mathrm{d}y}\\ &=2\int_{0}^{1}{\left(1-y\right)^{2}\left(\left(y+\frac{1}{2}\right)\int_{0}^{+\infty}{\frac{\sin^{2}{u}}{u^{2}}\,\mathrm{d}u}-\left\lvert y-\frac{1}{2}\right\rvert\int_{0}^{+\infty}{\frac{\sin^{2}{v}}{v^{2}}\,\mathrm{d}v}\right)\,\mathrm{d}y}\\ &=\pi\int_{0}^{1}{\left(1-y\right)^{2}\left(\left(y+\frac{1}{2}\right)-\left\lvert y-\frac{1}{2}\right\rvert\right)\mathrm{d}y}\\ &=\pi\int_{0}^{1}{\left(1-y\right)^{2}\left(y+\frac{1}{2}\right)\mathrm{d}y}-\pi\int_{0}^{\frac{1}{2}}{\left(1-y\right)^{2}\left(\frac{1}{2}-y\right)\mathrm{d}y}-\pi\int_{\frac{1}{2}}^{1}{\left(1-y\right)^{2}\left(y-\frac{1}{2}\right)\mathrm{d}y}\\ &=\frac{\pi}{4}-\frac{17\pi}{192}-\frac{\pi}{192}\\ &=\frac{5\pi}{32}\end{aligned}

Answer 3

A common way to evaluate these types of integrals is to use the Laplace/Mellin transform $$\int_{0}^{\infty} x^{\alpha-1} e^{-tx} \, \mathrm dx = \frac{\Gamma(\alpha)}{t^{\alpha}}, \quad \alpha, t >0. $$

$ \begin{align} \int_{0}^{\infty} \sin(t) \, \frac{\sin(t)-t}{t^{4}}\, \mathrm dt &= \frac{1}{\Gamma(4)} \int_{0}^{\infty} \sin(t) \left(\sin(t)-t \right) \int_{0}^{\infty} x^{3} e^{-tx} \, \mathrm dx \, \mathrm dt \\ &= \frac{1}{6} \int_{0}^{\infty} x^{3} \int_{0}^{\infty} \left(\sin^{2}(t) - t \sin(t) \right) e^{-xt} \, \mathrm dt \, \mathrm dx \\ &= \frac{1}{6} \int_{0}^{\infty} x^{3} \int_{0}^{\infty}\left(\frac{1- \cos(2t)}{2} - t \sin(t) \right) e^{-xt} \, \mathrm dt \, \mathrm dx \\ &\overset{\spadesuit}{=} \frac{1}{6} \int_{0}^{\infty} x^{3} \left(\frac{1}{2x} - \frac{1}{2} \frac{x}{x^{2}+4} - \frac{2x}{(1+x^{2})^{2}} \right) \, \mathrm dx \\ &\overset{\clubsuit}{=} \small \frac{1}{6} \int_{0}^{\infty} \left(\frac{x^{2}}{2} - \frac{1}{2} \left(x^{2}- 4 +\frac{16}{x^{2}+4} \right)-2 \left(1- \frac{2}{x^{2}+1}+ \frac{1}{(x^{2}+1)^{2}} \right) \right) \, \mathrm dx \\ &= \frac{1}{6} \int_{0}^{\infty} \left(-\frac{8}{x^{2}+4} + \frac{4}{x^{2}+1} - \frac{2}{(x^{2}+1)^{2}} \right) \, \mathrm dx \\ &= \frac{1}{6} \left(- 2 \pi + 2 \pi - \frac{\pi}{2} \right) \\ &= - \frac{\pi}{12} \end{align}$

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$\spadesuit$ $\int_{0}^{\infty} t \sin(\beta t) e^{-xt} \, \mathrm dt = -\frac{\mathrm d}{\mathrm d \beta} \int_{0}^{\infty} \cos(\beta t) e^{-xt} \, \mathrm dt$

$\clubsuit$ Polynomial long division

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For the second integral use the trig identity $\sin^{3}(t) = \frac{3\sin(t)-\sin(3t)}{4} $.

Answer 4

The case of $$I(a,b,c,d)=\int\frac{\sin^a (t)\left(\sin ^b(t)-t^c\right)}{t^d}\,dt$$ is workable since already given by formula $2.631.2$ in the seventh edition of the Table of Integrals, Series, and Products by I.S. Gradshteyn and I.M. Ryzhik. The formula is not too difficult to obtain using a couple of integration by parts.

The result is given in terms of sines, cosines and their corresponding integral functions such as $$I(1,2,3,4)=-\frac{\sin ^2(t) \cos (t)}{2 t^2}-\frac{\sin (t) \left(3 t^2+\left(9 t^2-2\right) \cos (2 t)+2\right)}{12 t^3}+$$ $$\frac{9 \,\text{Ci}(3 t)-\text{Ci}(t)}{8} -\text{Si}(t)$$

Around $t=0$, the integrand is $\sim t^{a-d} \left(t^b-t^c\right)$ and, if $b=c$ as in the post $$\frac{\sin^a (t)\left(\sin ^b(t)-t^c\right)}{t^d}=- t^{a+b-d}\,\left(\frac{b }{6}t^2+O\left(t^4\right)\right)$$

Interesting is the case of the definite integral $I(1,n,n,n+3)$ which is the generalization of your two integrals. They write $$I(1,n,n,n+3)=-a_n\, \frac \pi {12}$$

the first $a_n$ being $$\left\{1,\frac{15}{8},\frac{13}{5},\frac{623}{192},\frac{134}{35} ,\frac{67069}{15360},\frac{920}{189},\frac{9184703}{1720320},\frac{192431}{33264},\frac{11536764187}{1857945600}\right\}$$ which do not seem to correspond to any known sequence.