Integral of $\int_0^{\infty} \frac{\sin^2(x)}{x^2+1}dx$ using Feynman integration.

Question

Using $$I(t) = \int_0^\infty \frac{\sin^2(tx)}{x^2+1}dx$$ I want to know how to get an answer using Feynman integration and the Laplace transform of a differential equation. The correct answer is $\frac{(1-e^{-2})\pi}{4}$, but I keep getting $(1-e^{-2})\pi$, so I want to see where I have made a mistake.

Here is the method: setting $t = 1$ provides the integral in question. By repeatedly differentiating $I(t)$, you can obtain the differential equation that $4 I'(t) = I'''(t)$. Setting $J(t) = I'(t)$, use a Laplace transform to obtain $J(t)$. Now integrate $\int_0^1 J(t)dt$, which is equal to $I(1) - I(0)$ from the second fundamental theorem of calculus. Since $I(0) = 0$, solving for $I(1)$ yields the integral in question.

Keep in mind that I'm still in high school, so all I really know how to do is partial derivatives and Laplace transforms to solve differential equations. If something is beyond the subjects of multivariable calculus, please continue to answer the question, but know that what I am looking for is an answer through the Feynman technique and Laplace transforms.

Answer 1

First, note that $\sin^2(tx)=\frac12(1-\cos(2tx))$. Hence, we see that

$$I(t)=\frac\pi4-\frac12 \int_0^\infty \frac{\cos(2tx)}{x^2+1}\,dx\tag1$$

Differentiating under the integral in $(1)$ can be justified by noting that the integral $\int_0^\infty \frac{x\sin(2tx)}{x^2+1}\,dx$ converges uniformly for $|t|\ge \delta>0$. Proceeding reveals

$$\begin{align} I'(t)&=\int_0^\infty \frac{x\sin(2tx)}{x^2+1}\,dx\\\\ &=\int_0^\infty \frac{(x^2+1-1)\sin(2tx)}{x(x^2+1)}\,dx\\\\ &=\int_0^\infty \frac{\sin(2tx)}{x}\,dx-\int_0^\infty \frac{\sin(2tx)}{x(x^2+1)}\,dx\\\\ &=\frac\pi2 \text{sgn}(t)-\int_0^\infty \frac{\sin(2tx)}{x(x^2+1)}\,dx\tag2 \end{align}$$

Similarly, we can differentiate $(2)$ to obtain

$$\begin{align} I''(t)&=-2\int_0^\infty \frac{\cos(2tx)}{x^2+1}\,dx\\\\ &=4I(t)-\pi\tag3 \end{align}$$

From $(3)$ we have $I''(t)-4I(t)=-\pi$, while from $(1)$ we see that $I(0)=0$ and from $(2)$ we see that $\lim_{t\to 0^\pm}I'(t)=\pm \frac\pi2$. Solving this ODE with these initial conditions, we find

$$I(t)=\frac\pi4 -\frac\pi4 e^{-2|t|}$$

Answer 2

Hello felllow high school student :)

Here is the solution for $t\ge 0$. You can naturally continue it for any $t\in\mathbb{R}$.

\begin{align} I'(t)&=\int_0^\infty \frac{2\sin (tx) \cos(tx) x}{1+x^2}\,dx \\ &=\int_0^\infty \frac{x\sin(2xt)}{1+x^2}\,dx \\ &=\int_0^\infty \mathcal{L}^{-1}\left\{\frac{x}{1+x^2}\right\}(s)\cdot \mathcal{L}\left\{ \sin(2xt)\right\}(s)\,ds \qquad (1)\\ &=2t\int_0^\infty \frac{\cos(s)}{4t^2+s^2}\,ds \\ &=t \int_{-\infty}^\infty\frac{\cos(s)}{4t^2+s^2}\,ds \\ &=t\int_{-\infty}^\infty \cos(s)\int_0^\infty e^{-\nu(4t^2+s^2)}\,d\nu ds \\ &=t\int_0^\infty e^{-4t^2\nu} \int_{-\infty}^\infty \cos(s) e^{-\nu s^2}\,dsd\nu \\ &=t\int_0^\infty e^{-4t^2\nu} \int_{-\infty}^\infty e^{-\nu s^2+is}\,dsd\nu \\ &=t\int_0^\infty e^{-4t^2\nu-\frac{1}{4\nu}} \int_{-\infty}^\infty e^{-\nu \left(s+\frac{i}{2\nu}\right)^2}\,dsd\nu \\ &=\sqrt{\pi}t\int_{0}^\infty e^{-4t^2\nu-\frac{1}{4\nu}}\frac{d\nu}{\sqrt{\nu}} \qquad {\lambda=2\sqrt{\nu}} \\ &=\frac{\sqrt{\pi}}{2}t \int_{-\infty} ^\infty e^{-t^2\lambda^2-\frac{1}{\lambda^2}}\, d\lambda \\ &=t\frac{\sqrt{\pi}}{2} e^{-2t}\int_{-\infty} ^\infty e^{-t^2\left(\lambda-\frac{1} {t\lambda}\right)^2}\, d\lambda \qquad (2)\\ &=t\frac{\sqrt{\pi}}{2} e^{-2t}\int_{-\infty} ^\infty e^{-t^2\lambda^2}\, d\lambda \qquad (3) \\ &=\frac{\sqrt{\pi}}{2}t e^{-2t}\sqrt{\frac{\pi}{t^2}} \\ &=\frac{\pi}{2} e^{-2t} \end{align} Where I have used a property of the Laplace tranform in $(1)$, Glasser's Master Theorem in $(2)$, and the Gaussian integral in $(3)$. As we know that $I(0)=0$, we can integrate this equation from $0$ to $1$ to obtain

\begin{align} I(1)&=\frac{\pi}{2}\int_0^1 e^{-2t}\,dt \\ &=\frac{\pi}{4}\left(1-e^{-2}\right) \end{align}

Feel free to ask if you have any questions. I know there's a lot in this answer, but I think that you can definitely learn some nice tricks from it.