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.

  • $\begingroup$ Shouldn't the final answer depend on $t$? (I assume there is some typo). $\endgroup$
    – Ian
    Nov 14 '18 at 3:44
  • 1
    $\begingroup$ well... the "Feynman integration" was discovered some centuries ago by Leibniz :) $\endgroup$
    – Masacroso
    Nov 14 '18 at 3:45
  • $\begingroup$ @Masacroso The required theorem dates back to Leibniz, but I don't think the idea of writing an integral $I$ as $\frac{dJ}{da}(a_0)$ where $J$ is some other integral with a parameter $a$ in it really went back that far. $\endgroup$
    – Ian
    Nov 14 '18 at 3:46
  • 2
    $\begingroup$ Feynman just read the trick in a book and started to use it intensively $\endgroup$ Nov 14 '18 at 3:51
  • 1
    $\begingroup$ @Masacroso I simply called it Feynman integration as a term of generality. Not trying to diss papa Leibniz. $\endgroup$ Nov 14 '18 at 3:58

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

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

  • $\begingroup$ Thank you. I see now that my error occurred in finding the differential equation . $\endgroup$ Nov 14 '18 at 5:14
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$
    – Mark Viola
    Nov 14 '18 at 13:27

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.

  • $\begingroup$ Thank you for your answer. Quite honestly I have no idea whats going on here at the moment, but that gets me pretty excited to learn about it later. That Gaussian integral seems to pop up everywhere. Very cool solution. $\endgroup$ Nov 14 '18 at 5:22
  • $\begingroup$ I like your excitement! $\endgroup$
    – Zachary
    Nov 14 '18 at 5:23
  • $\begingroup$ @SuchetanDontha - I'm going through a bit of a Feynman + Laplace phase at the moment. Here's my recent ones (thought you may be interested) math.stackexchange.com/search?q=user:150203+[laplace-transform] $\endgroup$
    – user150203
    Nov 21 '18 at 5:24

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