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I want to find the expression for the following integral

$$\int_0^\infty\text{d}x\frac{e^{i k x}}{x}$$

I have tried deriving with respect to $k$, transforming into an integral over the whole real line... with no luck. Would anyone give me a hint or the actual steps to take? Thank you so much.

EDIT: Following @Dr. MV's advice I used the Convolution Theorem $\mathcal{F}[f\cdot g]=\mathcal{F}[f]*\mathcal{F}[g]$ to arrive to the following expression

$$\mathcal{F}\left[\frac{1}{x}\Theta(x)\right]=-i\pi^2\text{sign}(k)+\pi\int_{-\infty}^{\infty}\text{d}t\frac{\text{sign}(t)}{t-k}$$

So now the problem is this last integral, which diverges in both extremes and in $t=k$. Interpreting $\text{sign}(t)$ as a distribution, I can perform the change of variables $y=t-k$, which gives me

$$f(k)=\int_{-\infty}^{\infty}\text{d}y\frac{\text{sign}(y+k)}{y}$$

Now deriving with respect to $k$ gives me

$$\frac{df(k)}{dk}=2\int_{-\infty}^\infty\text{d}y\frac{\delta(y+k)}{y}=\frac{2}{-k}$$

And now integrating gives me $f(k)=-2\log(k)$, so

$$\mathcal{F}\left[\frac{1}{x}\Theta(x)\right]=-i\pi^2\text{sign}(k)-2\pi\log(k)$$

My question now is, is this reasoning valid for all values of $k$? In particular, I am thinking of $k<0$.

Alex
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  • The integral diverges for all $k$. – Mark Viola Feb 08 '16 at 17:37
  • Are you sure of the lower limit? You may be trying to determine the Fourier Transform of $\frac{1}{x}$. – jim Feb 08 '16 at 17:40
  • Yes, I'm quite sure of the lower limit, the integral actually comes from a physical problem. In terms of Fourier transforms, I also tried computing $\mathcal{F}\left[\frac{1}{x}\Theta(x)\right]$, where $\Theta(x)$ is the Heaviside step function. – Alex Feb 08 '16 at 20:18
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    @pepeperez While the integral diverges, the notation can be interpreted as the Fourier Transform that you tried computing. Use the convolution theorem and the Fourier Transforms for $1/x$ and the Heaviside function and see if this provides a way forward. - Mark – Mark Viola Feb 08 '16 at 20:25

1 Answers1

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Actually I have found another solution, more satisfactory to my taste (although maybe not 100% complete). Inspired by the answer of this question on integrating the $\text{sinc}$ function, I thought of the following procedure:

  1. Write the exponential as a sum of sines and cosines

$$\int_0^\infty\text{d}x\frac{e^{i k x}}{x}=\int_0^\infty\text{d}x\frac{\cos(k x)}{x}+i\int_0^\infty\text{d}x\frac{\sin(k x)}{x}$$

  1. Write $\frac{1}{x}=\int_0^\infty e^{-x t}\text{d}t$ and perform the Laplace Transforms

$$\begin{align} \int_0^\infty\text{d}x\frac{e^{i k x}}{x}&=\int_0^\infty\text{d}t\int_0^\infty\text{d}x\cos(k x)e^{-x t}+i\int_0^\infty\text{d}t\int_0^\infty\text{d}x\sin(k x)e^{-x t}\\ &=\int_0^\infty\text{d}t\frac{t}{k^2+t^2}+i\int_0^\infty\text{d}t\frac{k}{k^2+t^2} \end{align}$$

  1. Perform the remaining integrations

$$\begin{align} \int_0^\infty\text{d}x\frac{e^{i k x}}{x}=&\left.\frac{1}{2}\log(k^2+t^2)\right|_0^\infty+i\arctan\left.\left(\frac{t}{k}\right)\right|_0^\infty\\ =&\left.\frac{1}{2}\log(k^2+t^2)\right|_{t\rightarrow\infty}-\log|k|+i\frac{\pi}{2}\text{sign}(k) \end{align}$$

Where the divergent term is some constant independent of $k$ which I have intentionally left there (see reason below). Computing the same integral with Mathematica gives

FourierTransform[1/x HeavisideTheta[x], x, k, FourierParameters -> {1, 1}] (* 1/2 (-2 EulerGamma - 2 Log[Abs[k]] + I Pi Sign[k]) *)

Which coincides with the answer given up to the Euler-Mascheroni constant, which I assume comes from the divergent part when treating it accordingly.

Thank you all for your help and tips.

Community
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Alex
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