Initially this integral looked familiar but soon I noticed the $x^2$. I tried to solve it however I could before quickly (albeit very quickly) moving onto online solvers and Mathematica.

The online solves can't give a solution and Mathematica also doesn't.

Here:$$\frac{1}{2\pi\beta}\int_{-\infty}^{\infty}\frac{1}{k^2}e^{ikx}\sin(\beta tk^2)dk$$

  • 2
    $\begingroup$ Can you find the Fourier transform of $\sin (x^2)$? If so, you could try differentiating by $t$. $\endgroup$
    – user27182
    Dec 31, 2021 at 12:23
  • $\begingroup$ Mathematica can give solution try: FourierTransform[1/(2 Pi \[Beta])*Sin[\[Beta] t k^2]/k^2, k, x, FourierParameters -> {1, -1}] $\endgroup$ Jan 1 at 12:51

1 Answer 1


hope you're well!

Using differentiation under the integral sign, one can parametrize the given integral as $$ I(a) = \frac{1}{2\pi\beta}\int^{+\infty}_{-\infty}\frac{e^{ikx}}{k^2}\sin\left(a\beta tk^2\right) \mathrm{d}k \Rightarrow \frac{\mathrm{d}I(a)}{\mathrm{d}a} = \frac{t}{2\pi} \int^{+\infty}_{-\infty}e^{ikx}\cos\left(a\beta tk^2\right) \mathrm{d}k. $$

Since $\cos\alpha = (e^{i\alpha} + e^{-i\alpha})/2$, we can split the derivative into two integrals: $$ \frac{\mathrm{d}I(a)}{\mathrm{d}a} = \frac{t}{4\pi}\int^{+\infty}_{-\infty} \left[e^{ik(x+a\beta tk)} + e^{ik(x-a\beta tk)}\right] \mathrm{d}k. $$

Complete the square of both exponents. Notice that $$ \exp\left[ik(x\pm a\beta tk)\right] = \exp\left[\pm i\left(\sqrt{a\beta t}k \pm \frac{x}{2\sqrt{a\beta t}}\right)^2\right]\exp\left(\mp i\frac{x^2}{4a\beta t}\right). $$ Taking $s = \sqrt{a\beta t}k \pm x/2\sqrt{a\beta t}$ respectively, we have that $$ \begin{aligned} \int^{+\infty}_{-\infty} e^{ik(x+a\beta tk)} \mathrm{d} k &= \frac{\exp\left(- i\frac{x^2}{4a\beta t}\right)}{\sqrt{a\beta t}} \int^{+\infty}_{-\infty} e^{is^2}\mathrm{d}s = \frac{\exp\left(- i\frac{x^2}{4a\beta t}\right)}{\sqrt{a\beta t}} (1+i)\sqrt{\frac{\pi}{2}}, \\ \int^{+\infty}_{-\infty} e^{ik(x-a\beta tk)} \mathrm{d} k &= \frac{\exp\left( i\frac{x^2}{4a\beta t}\right)}{\sqrt{a\beta t}} \int^{+\infty}_{-\infty} e^{-is^2}\mathrm{d}s = \frac{\exp\left(i\frac{x^2}{4a\beta t}\right)}{\sqrt{a\beta t}} (1-i)\sqrt{\frac{\pi}{2}}. \end{aligned} $$

Thus, $$ \begin{aligned} \frac{\mathrm{d}I(a)}{\mathrm{d}a} &= \frac{t}{4\pi}\frac{1}{\sqrt{a\beta t}}\sqrt{\frac{\pi}{2}}\left[(1+i)\exp\left(- i\frac{x^2}{4a\beta t}\right) + (1-i)\exp\left(i\frac{x^2}{4a\beta t}\right)\right] \\ &= \frac{1}{2\sqrt{2\pi}}\sqrt{\frac{t}{a\beta}}\left(\cos\frac{x^2}{4a\beta t} + \sin\frac{x^2}{4a\beta t}\right). \end{aligned} $$

Now, we integrate with respect to $a$. Following the output of WolframAlpha, $$ \begin{aligned} I(a) &= \frac{1}{\sqrt{2\pi}} \sqrt{\frac{t}{\beta}} \left[ \sqrt{2\pi} \sqrt{\frac{x^2}{4\beta t}} \left( \mathcal{S}\left(\sqrt{\frac{2x^2}{4\pi a\beta t}}\right) - \mathcal{C}\left(\sqrt{\frac{2x^2}{4\pi a\beta t}}\right) \right) + \sqrt{a}\left(\cos\frac{x^2}{4a\beta t} + \sin\frac{x^2}{4a\beta t}\right) \right] + C \\ &= \frac{1}{\sqrt{2\pi}} \sqrt{\frac{t}{\beta}} \left[ \sqrt{\frac{\pi x^2}{2\beta t}} \left( \mathcal{S}\left(\sqrt{\frac{x^2}{2\pi a\beta t}}\right) - \mathcal{C}\left(\sqrt{\frac{x^2}{2\pi a\beta t}}\right) \right) + \sqrt{a}\left(\cos\frac{x^2}{4a\beta t} + \sin\frac{x^2}{4a\beta t}\right) \right] + C, \end{aligned} $$ where $\mathcal{C}(x)$ and $\mathcal{S}(x)$ are the cosine and sine Fresnel integrals defined as $$ \mathcal{C}(x) = \int^x_0 \cos\frac{\pi t^2}{2} \mathrm{d}t, \qquad \mathcal{S}(x) = \int^x_0 \sin\frac{\pi t^2}{2} \mathrm{d}t. $$

Notice that $I(a=0) = 0$ is satisfied only if $C = 0$ since $$ \int^{+\infty}_0 \cos\frac{\pi t^2}{2} \mathrm{d}t = \int^{+\infty}_0 \sin\frac{\pi t^2}{2} \mathrm{d}t = \frac{1}{2}. $$

Therefore, we can retrieve the desired integral setting $a=1$: $$ I(a=1) = \frac{1}{\sqrt{2\pi}} \sqrt{\frac{t}{\beta}} \left[ \sqrt{\frac{\pi x^2}{2\beta t}} \left( \mathcal{S}\left(\sqrt{\frac{x^2}{2\pi \beta t}}\right) - \mathcal{C}\left(\sqrt{\frac{x^2}{2\pi \beta t}}\right) \right) + \cos\frac{x^2}{4\beta t} + \sin\frac{x^2}{4\beta t} \right]. $$

For $(x, \beta, t) = (0, 1, 1)$, the result works as expected. I have also checked the $(1, 1, 1)$ case. Hope you like it!

  • 1
    $\begingroup$ You are missing some parentheses for the Fresnel integrals. $\endgroup$
    – Gary
    Jan 1 at 4:45
  • 1
    $\begingroup$ Corrected! @Gary $\endgroup$ Jan 1 at 4:57

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