Using Fourier’s transform to solve integral $ \int_{-\infty }^{\infty }\frac{t^2\cos^2t+\sin^2t-t\sin2t}{t^4}dt $

Question

How do we use Fourier’s transform to solve integral:

$$ \int_{-\infty }^{\infty }\frac{t^2\cos^2(t)+\sin^2(t)-t\sin(2t)}{t^4}dt $$

The answer is $\pi/3.$

May I know the approach for solving this integral? I think Parseval’s theorem is one of the approach. Apart from that, is there any other methods to solve this integral?

Answer 1

If the complex integration is acceptable, we can do the following. Using the relations $\displaystyle \cos^2t=\frac{1+\cos2t}{2}; \,\sin^2t=\frac{1-\cos2t}{2}$ and making the substitution $t=\frac{x}{2}$, we can write the integral in the form $$I=\int_{-\infty}^\infty\Big(x^2(1+\cos x)+4(1-\cos x)-4x\sin x\Big)\frac{dx}{x^4}$$ $$=\Re\int_{-\infty}^\infty\Big(x^2(1+e^{ix})+4(1-e^{ix})+4ixe^{ix}\Big)\frac{dx}{x^4}=\Re\int_{-\infty}^\infty f(x)dx=\Re J$$ Now we create a closed contour, adding a big half-circle in the upper half of the complex plane ($I_R$, counter-clockwise) and a small half-circle around $x=0$ ($I_r$, clockwise). There are no singularities inside our contour; therefore $$\oint f(x)dx=J+I_R+I_r=0$$ Integral along a big half-circle $I_R\,\to 0$ as $R\to\infty$ , and we are left with $$J=-I_r=\pi i\,Res_{x=0}f(x)$$ To find the residue, we decompose the integrand near $x=0$: $$f(x)=\frac{1}{x^4}\Big(x^2(2+ix+...)+4(-ix+\frac{x^2}{2!}+\frac{ix^3}{3!}+..)+4ix(1+ix-\frac{x^2}{2!}+..)\Big)$$ The residue is the coefficient at $\displaystyle\frac{1}{x}:\,\,\,i+\frac{4}{6}i-2i=-\frac{i}{3}$ $$I=\Re\,J=\Re\,\Big(\pi i\cdot\frac{-i}{3}\Big)=\frac{\pi}{3}$$

Answer 2

Use $\sin(2t) = 2\sin(t) \cos(t)$. Then the numerator becomes $(t\cos(t) - \sin(t))^2$ so the integral is $$\int^\infty_{-\infty} \left(\frac{t\cos(t)-\sin(t)}{t^2}\right)^2dt.$$ As you have pointed out, you can now use Parseval's theorem if you can find a function whose Fourier transform is (a constant multiple of) $$\hat f(\omega) = \frac{\omega \cos(\omega) - \sin(\omega)}{\omega^2}.$$ Try something like $$f(x) = \begin{cases}x, & \vert x \vert \le 1, \\ 0, & \text{otherwise} \end{cases}$$

Answer 3

Another approach via integration by parts\begin{align} &\int_{-\infty }^{\infty }\frac{t^2\cos^2t+\sin^2t-t\sin2t}{t^4}dt\\ =&\ \frac13 \int_{-\infty }^{\infty }\left(t^2\cos^2t+\sin^2t-t\sin2t\right)d\left( -\frac1{t^3}\right)\\ \overset{ibp}=& \ \frac13\int_{-\infty }^{\infty }\frac{1-\cos2t}{t^2}\ \overset{ibp} {dt} - \frac13 \int_{-\infty }^{\infty }\frac{\sin2t}{t}dt\\ =& \ \frac13 \int_{-\infty }^{\infty }\frac{\sin2t}{t} \overset{2t\to t}{dt}= \frac13 \int_{-\infty }^{\infty } \frac{\sin t}{t}dt=\frac\pi3 \end{align} where $\int_0^\infty \frac{\sin t}t dt = \int_0^\infty \int_0^\infty {\sin t}\ e^{-tu}dudt = \int_0^\infty\frac1{1+u^2}du = \frac\pi2$