Calculate $\int_{-\infty}^{\infty} \frac{\sin(t)}{t(1+t^2)} dt$

Question

$$\int_{-\infty}^{\infty} \frac{\sin(t)}{t(1+t^2)} dt$$

$$\int_{-\infty}^{\infty} \frac{\sin(t)}{t(1+t^2)} dt={\bf Im} \int_{-\infty}^{\infty} \frac{e^{it}}{t(1+t^2)} dt $$ for denominator roots $$t(1+t^2)=0$$ then roots are: $$(0,i,-i)$$

I am going to calculate the integral value by determining residues associated to $0$ and $i$

\begin{eqnarray} {\bf Im}\int_{-\infty}^{\infty} \frac{e^{it}}{t(1+t^2)} dt &=&{\bf Im}\Big( \pi i \ \lim_{t \to 0} \frac{e^{it}}{1+t^2}+2 \pi i \ \lim_{t \to i} \frac{e^{it}}{t(t+i)}\Big)\\&=&{\bf Im}(\pi i + 2 \pi i \frac{e^{-1}}{-2})\\&=&\pi(1-e^{-1}) \end{eqnarray}

Is it correct?

How can I apply Plancherel theorem to calculate the same integral?

Thanks!

Answer 1

Since $\mathcal{L}(\sin t)=\frac{1}{1+s^2}$ and $\mathcal{L}^{-1}\left(\frac{1}{t(t^2+1)}\right)=1-\cos(s)$, the value of the integral is $$ 2\int_{0}^{+\infty}\frac{1-\cos(s)}{1+s^2}\,ds = \pi-2\int_{0}^{+\infty}\frac{\cos(s)}{1+s^2} =\color{red}{\pi\left(1-\frac{1}{e}\right).}$$

Answer 2

I verified your result using

\begin{align} \int_{-\infty}^{\infty} \frac{\sin(t)}{t(1+t^2)} dt &= 2\int^\infty_0 \frac{t\sin(t)}{t^2(1+t^2)}\,dt\\ & = 2\int^\infty_0 \frac{\sin(t)}{t}\,dt - 2\int^\infty_0 \frac{t\sin(t)}{1+t^2} \,dt\\&= \pi -\frac{\pi}{e} \end{align}

Both integrals are standards in contour integration.

Answer 3

It seems no one has yet adressed the question of how to do this using Plancherel's theorem. Here it goes:

The Fourier transform of $f(x) := \chi_{[-1,1]}(x)$ is easily computed to be $$\hat f (t) = \sqrt{\frac{2}{\pi}} \frac{\sin(t)}{t},$$ and the Fourier transform of $g(x) := \exp(-|x|)$ is $$\hat g(t) =\frac{1}{\sqrt{2\pi}} \frac{2}{1+t^2}$$ (See this question for a proof of the latter). Using Plancherel's theorem, we obtain

$$\int_{-\infty}^{\infty} \frac{\sin(t)}{t(1+t^2)}\mathrm{d}t = \frac{\pi}{2} \cdot (\hat f, \hat g)_{L^2} = \frac{\pi}{2}\cdot (f,g)_{L^2} = \pi \int_{0}^{1} e^{-x} ~\mathrm{d}x = \pi(1-e^{-1}) $$

Answer 4

Let us consider $$I\left(x\right)=2\int_{0}^{\infty}\frac{\sin\left(xt\right)}{t\left(1+t^{2}\right)}dt$$ then $$I'\left(x\right)=2\int_{0}^{\infty}\frac{\cos\left(xt\right)}{1+t^{2}}dt$$ and following this proof we get $$I'\left(x\right)=\pi e^{-x} $$ then $$I\left(x\right)=-\pi e^{-x}+C $$ and since $I\left(0\right)=0 $ we have $$C=\pi $$ so $$I\left(1\right)=2\int_{0}^{\infty}\frac{\sin\left(t\right)}{t\left(1+t^{2}\right)}dt=\color{blue}{\pi-\frac{\pi}{e}}.$$