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I would like to compute the Fourier transform of the product of $\tanh(x)$ and the Heaviside step function $H(x)$, i.e.

$$\int_{-\infty}^{\infty} H(x)\tanh(x)e^{-ikx}dx = \int_{0}^{\infty} \tanh(x)e^{-ikx}dx$$

For the Fourier transform of $\tanh(x)$ alone, I have read that the use of differentiation:

$$ik\mathcal{F}\left[ \tanh(x)\right](k) = \mathcal{F}\left[ \text{sech}^2(x)\right](k)$$

is a possible way of proceding, as the integral does not exist as a classical Riemann integral. I have then read derivations of the Fourier transform of $\text{sech}^2(x)$ using contour integration.

Adapting for the new problem I have supposed:

$$ik\mathcal{F}\left[ H(x)\tanh(x)\right](k) = \mathcal{F}\left[H(x) \text{sech}^2(x)\right](k)$$

However, I cannot work out how to adapt contour integration to the half line integral from 0 to $\infty$ instead.

Can someone explain how to compute the Fourier transform either by the method outlined above or a different method?

Thank you!

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  • $\begingroup$ You cant try to add and subtract $\frac{1}{2} sech^2 (x)$ inside the fourier transform. Then you can pull out terms that look more like the Fourier transform of $sech^2$ and a Hilbert transform. $\endgroup$
    – AHusain
    Apr 23, 2020 at 23:01
  • $\begingroup$ The regular part is $$\int_{\mathbb R} (\tanh(x) H(x) - H(x)) \, e^{-i k x} dx = \frac 1 2 \psi {\left( \frac {i k} 4 + \frac 1 2 \right)} - \frac 1 2 \psi {\left( \frac {i k} 4 + 1 \right)},$$ which leaves the singular part $\mathcal F[H](k)$. $\endgroup$
    – Maxim
    Apr 24, 2020 at 7:10
  • $\begingroup$ What is $\psi$ here? $\endgroup$
    – mshaw
    Apr 28, 2020 at 20:15
  • $\begingroup$ $\psi$ is the polygamma function. $\endgroup$
    – Maxim
    May 6, 2020 at 20:50

1 Answer 1

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notice that: $$I=\int_{-\infty}^\infty \text{H}(x)\tanh(x)e^{-ikx}dx=\int_0^\infty\frac{e^{2x}-1}{e^{2x}+1}e^{-ikx}dx=\int_0^\infty\frac{u^{-ik}(u-1)}{u^2+1}$$ now notice that: $$I=\int_0^\infty\frac{u^{1-ik}}{u^2+1}-\frac{u^{-ik}}{u^2+1}du$$ if we let: $$J(a)=\int_0^\infty\frac{u^a}{u^2+1}du=\frac{1}{2}\int_1^\infty(v-1)^{\frac{a-1}{2}}v^{-1}dv=\frac{(-1)^{\frac{a-1}{2}}}{2}\int_1^\infty(1-v)^{\frac{a-1}{2}}v^{-1}dv$$ $$J(a)=\lim_{z\to\infty}\frac{(-1)^{\frac{a-1}{2}}}{2}\left[B\left(z;0,\frac{a+1}{2}\right)-B\left(1;0,\frac{a+1}{2}\right)\right]$$ and we know that: $$I=J(1-ik)-J(-ik)$$ where $B$ is the incomplete beta function

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  • $\begingroup$ Shouldn't it read $\int_0^\infty\frac{u^{-ik}(u-u^{-1})}{u^2+1}du$ at the end of the first line? $\endgroup$
    – mshaw
    Apr 28, 2020 at 13:41

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