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I have the task to evaluate $|a(k)|^2$ with $$ a(k) = \int_{-\infty}^\infty \!dx\,\exp\left(i k x + i e^{x}\right).\tag{1}$$

The integral in (1) can be evaluated explicitly via the substitution $y=e^{x}$ with the result $$ a(k) = e^{-\pi k/2} \Gamma(ik)$$ where $\Gamma(x)$ is the Gamma-function. Using the result $$\Gamma(ix)\Gamma(-ix) = \frac{\pi}{x \sinh(\pi x)} \tag{2}$$ we find $$|a(k)|^2 = \frac{2\pi}{k(e^{2\pi k} -1)}.$$

Does anybody know a direct way to evaluate $|a(k)|^2$ possibly using some methods of complex analysis such that one does not need to know the special property (2) of the Gamma-function (and even the Gamma-function itself)?

Edit1: (see also edit 2 about the question of convergence) There was a question about possible problems with the convergence of the integral. Let me indicate in what sense one can give a meaning to the formulas above. Eq. (1) converges in the half-plane $\text{Im}(k) < 0$. Of course, we want in the end $k \in \mathbb{R}$. We should understand this as a limes $$a(k)=\lim_{\eta\downarrow 0}\int_{-\infty}^\infty \!dx\,\exp\left(i k x - \eta x + i e^{x}\right)$$ and we ask for the value of $$ |a(k)|^2= \lim_{\eta\downarrow 0}\int_{-\infty}^\infty \!dx\,dy\,\exp\left(i k x - \eta x + i e^{x} -i k y - \eta y - i e^{y}\right) $$

To show that (1) is indeed convergent for $ \text{Im}(k) < 0$, we employ the substitution $y= e^x$ and obtain $$a(k)= \int_0^\infty\!dy\,e^{i y} y^{i k -1}.$$ The magnitude of the integrand is $\sim \exp[-\text{Im}(y)]$ for $|y|\to \infty$ and the integrand has no singularities. So we can deform the integration contour along the positive imaginary axis with $y=iz$ and $$a(k) = i\int_0^\infty\!dz\, e^{-z} (iz)^{ik-1} $$ which is convergent for $\text{Re}(ik -1)= -\text{Im}(k) -1 >-1$.

Edit2: (about the convergence)

As the question about the convergence of the integral as defined arises again and again, here is some explanation without using complex analysis (which might be simpler to understand for some). Note that as before, we understand $a(k)$ as $\lim_{\eta\downarrow 0}a(k-i \eta)$.

Starting with Eq. (1) we employ the change of variables $y=e^{x}$. The limits of the resulting integral are $y\in[0,\infty)$ which we divide in two integrals which we will analyze individually, $$a(k) = a_1(k) + a_2(k) = \int_0^1\!dy\, y^{ik-1} e^{iy} + \int_1^\infty\!dy\, y^{ik-1} e^{iy}.$$

It is only for the part $a_1(k)$ where we need to care about the limit $\eta\downarrow 0$. In particular, we can perform an integration by parts (integrating $y^{ik-1}$) and obtain $$a_1(k) = -i k^{-1} e^{i} + k^{-1}\lim_{y\to 0} (i e^{iy} y^{ik}) - k^{-1}\int_0^1\!dy\, y^{ik} e^{iy} =-i k^{-1} e^{i}- k^{-1}\int_0^1\!dy\, y^{ik} e^{iy} .$$ For the vanishing of the limit, we have used the replacement $k\mapsto k-i\eta$ with $\eta \downarrow 0$. The rest is finite as $| y^{ik} e^{iy}| =1$.

The second part $a_2(k)$ is finite even for real $k$ (understanding the integral as an improper integral). To see that explicitly, we introduce $z(y)= y + k \log y$. For $y> y^* =\text{max}(1,-k)$ the variable change $z(y)$ is monotonously increasing. So we can write $$a_2(k) = \int_1^{y^*}\!dy\, y^{ik-1} e^{iy} + \underbrace{\int_{z^*}^\infty\!dz\, \frac{e^{iz}}{y(1+k/y)}}_{a_3(k)}.$$ The first term is trivially finite. The second term, we split in real and imaginary part. We show that the imaginary part is finite, the real part is analogous. We split the integral in parts with $z \in [ m \pi , (m+1) \pi]$ with $b_m= \int_{m\pi}^{(m+1)\pi}\!dz\,\sin(z)/y(1+k/y)$. We set $m^* = \lceil z^*/\pi\rceil$. We have $$\text{Im}\,a_3(k) = \int_{z^*}^{m^* \pi} \!dz\,\frac{\sin z}{y(1+k/y)} + \sum_{m=m^*}^\infty b_m.\tag{3}$$ We have that $(-1)^m b_m \geq 0$ thus $b_m$ is alternating in sign. Moreover, $$|b_m| \leq \frac{1}{m(1+k/m\pi)} \leq \frac1m \to 0 \quad (m\to\infty),$$ so the series in (3) converges due to the Leibniz criterion.

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  • $\begingroup$ I'm not very familiar with these constructions, but I wanted to ask: in what sense do we say the above integral converges? I ask because I can think of a few methods which would indicate the opposite result (the integrand isn't $L^1$ or $L^2$, nor would partitions of unity work in this case). $\endgroup$ – Jonathan Y. Jan 21 '14 at 9:15
  • $\begingroup$ @JonathanY.: this is of course a very relevant question. It is true that the integral only converges in a very weak sense. I will add some lines explaining that. $\endgroup$ – Fabian Jan 21 '14 at 9:32
  • $\begingroup$ Are we dealing with a Laplace transform, then (complex $k$)? Also, please excuse me, but I don't quite follow the proof of convergence (as the transform is two-sided). Namely, I get lost where you switched curves for integration. $\endgroup$ – Jonathan Y. Jan 21 '14 at 10:54
  • $\begingroup$ @JonathanY.: the transform is two-sided, however the substitution $y=e^{x}$ maps it onto a one sided interval. So I guess in the end you can see it as a Mellin-transform. $\endgroup$ – Fabian Jan 21 '14 at 12:43
  • $\begingroup$ Do you know of a good derivation for $(2)$ above? Does it involve the product of two integral expressions for $\Gamma$? If so, I think I see a path forward... $\endgroup$ – rajb245 May 8 '14 at 19:44
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this is how I'd show it. for $Re(s) > 0$ :

$$\Gamma(s) = \int_0^{+\infty} x^{s-1} e^{-x} dx$$

  • $z^{s-1} e^{-z}$ is holomorphic for $Re(z) > 0$ and decreases exponentially fast on every ray of this half plane

  • thus for any $a \in \mathbb{C},Re(a) >0$ : $\displaystyle\int_0^{+\infty} + \int_{+\infty}^{+a \infty}+\int_{+a\infty}^{0}z^{s-1} e^{-z}dz = 0$ and $$\Gamma(s) = \int_0^{+\infty} x^{s-1} e^{-x} dx = \int_0^{+a\infty} z^{s-1} e^{-z} dz$$

  • note that $\int_0^\infty y^{s-1} e^{iy} dy$ converges if $0 < Re(s) < 1$ so in that case the latter formula extends to $a = -i$ and : $$\Gamma(s) = \int_0^{-i\infty} z^{s-1} e^{-z} dz = \int_0^{+\infty} (-iy)^{s-1} e^{iy} d(-iy) = e^{i \pi s /2} \int_0^{+\infty} y^{s-1} e^{iy} dy = e^{i \pi s /2} \int_{-\infty}^{+\infty} e^{s u + i e^u} du$$

hence if $0 < Re(ik) < 1$ :

$$\int_{-\infty}^\infty exp\left(i k x + i e^{x}\right) dx= e^{- \pi k /2} \Gamma(ik)$$

and a simple analytic continuation argument shows that for $Re(ik) = 0$ if $\Gamma(ik)$ exists i.e. for $k \ne 0$ the last formula is still valid in the sense that $$ e^{- \pi k /2} \Gamma(ik) \ \ \text{ is the Fourier transform of } \ \ f(x) = \exp(i e^x)$$

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Are you sure your formula is correct? (consider what happens at $k=-1$). What you are looking for is the shifted Fourier transform $F(k+1)$ where $F$ is the Fourier transform of the constant function $1$. It is generally accepted that this is $2\pi\delta(x)$ (the Dirac delta function). This is justified by physicists using ad hoc empirical reasoning but can be demonstrated rigorously as a parametric interval in the sense of distribution theory. This was done by J. Sebastião e Silva in a seminal, but hard to find, article "Integrals and orders of growth of distributions" (Lisbon, 1964).

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  • $\begingroup$ If I set $k=-1$, my formula gives $a(-1)=e^{\pi/2} \Gamma(-i)\approx -0.74 + 2.39 i$, which is supported by numerics. I don't understand where the Dirac distribution comes into the game. Did you read the last part of my post, where I explain that I am interested in the limit $k \to -1$ with $\text{Im} (k) < 0$? Of course you could also try to give the integral a meaning via distribution theory but that does not look simpler to me (and will lead to the same result). $\endgroup$ – Fabian May 5 '14 at 7:54
  • $\begingroup$ I added some more comments using only real analysis, why the integral converges for $\text{Im}(k) <0$. $\endgroup$ – Fabian May 5 '14 at 9:48
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Mathematica 9.0 gave:

  Integrate[Exp[I k x + I Exp[x]], {x, -Infinity, Infinity}]

  ConditionalExpression[E^(-(k Pi/2) Gamma[I k], -1 < Im[k] < 0]

So the integral is convergent only when $-1<\text{Im}(k)<0$.

To understand why it is so, we break the original integral into two parts: $$a(k)=a_1(k)+a_2(k)$$

$$a_1(k)=\int_{-\infty}^0 \!dx\,\exp\left(i k x + i e^{x}\right)=-\int_0^{\infty} \!dy\,\exp\left(-i k y + i e^{-y}\right)$$

$$a_2(k)=\lim\int_{0}^\infty \!dx\,\exp\left(i k x + i e^{x}\right)$$

We can see that when $k=2-2i$ (i.e. $\text{Im}(k)<-1$), we have:

$$\exp(i k x)=\exp(i (2-2i) x)=\exp(i 2 x)\exp(2 x)$$

This factor in $a_2$ will be divergent.

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  • $\begingroup$ How do you know that you can break the integral up in two parts, that is to say how can you conclude from the divergence of the individual parts that the original integral did diverge? $\endgroup$ – Fabian Dec 12 '14 at 19:25
  • $\begingroup$ Indeed, $\int_{-\infty}^\infty x \, dx$ and $\int_{-\infty}^\infty -x \, dx$ diverge. But, $\int_{-\infty}^\infty x \, dx + \int_{-\infty}^\infty -x \, dx = \int_{-\infty}^\infty 0 \, dx = 0.$ $\endgroup$ – MathMajor Jun 29 '15 at 4:33

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