2
$\begingroup$

I'm asked to calculate the following integral for which $0 \neq a \in \mathbb{R}$:

$$\int_0^{\infty}\frac{\ln x}{\sqrt{x}(x^2+a^2)^2}$$

I'm confused about which contour I should use, whether it should be a semi-circle deformed to avoid the origin, or a keyhole contour based on a similar question I found here: Calculating $\int_0^{\infty } \frac{\ln (x)}{\sqrt{x} \left(a^2+x^2\right)^2} \, \mathrm{d}x$ using contour integration (but this question did not go into detail about how the integration was carried out using the keyhole contour)

Also, after deciding on which contour to use, how do I proceed from there to evaluate this integral? I know that I would eventually have to use the Residue Theorem but how do I isolate the part of the contour only going from $0$ to $\infty$?

$\endgroup$
  • $\begingroup$ Why the deliberate duplicate? $\endgroup$ – Did Nov 19 '17 at 10:07
  • $\begingroup$ The question I linked to didn't address how to use the keyhole contour to evaluate the integral; it focused on using a different contour $\endgroup$ – D. Ashton Nov 19 '17 at 10:23
1
$\begingroup$

We integrate with $a$ a positive real

$$f(z) = \frac{\mathrm{Log}(z)}{(z-ai)^2 (z+ai)^2} \exp(-1/2\times \mathrm{Log}(z))$$

around a keyhole contour with the slot on the positive real axis, which is also where the branch cut of the logarithm is located (argument of the logarithm is between $0$ and $2\pi$). Now for the large circle we get $\lim_{R\to\infty} 2\pi R \log R / \sqrt{R} /R^4 = 0$ so there is no contribution in the limit. For the small circle around the origin we find $\lim_{\epsilon\to\ 0} 2\pi \epsilon \log \epsilon / \sqrt\epsilon /a^4 = 0$ so there is no contribution here either.

We get for the upper line segment

$$\int_0^\infty \frac{\log x}{(x^2+a^2)^2} \exp(-1/2\times \log x) \; dx$$

which is our target integral, call it $J$. The lower line segment contributes

$$-\int_0^\infty \frac{\log x + 2\pi i}{(x^2+a^2)^2} \exp(-1/2\times \log x) \exp(-1/2\times 2\pi i) \; dx \\ = \int_0^\infty \frac{\log x + 2\pi i}{(x^2+a^2)^2} \exp(-1/2\times \log x) \; dx \\ = J + 2\pi i \int_0^\infty \frac{1}{(x^2+a^2)^2} \exp(-1/2\times \log x) \; dx = J + 2\pi i K$$

where $J$ and $K$ are real numbers. We thus have

$$2J + 2\pi i K = 2\pi i \mathrm{Res}_{z=ai} f(z) + 2\pi i \mathrm{Res}_{z=-ai} f(z)$$

or

$$J + \pi i K = \pi i \mathrm{Res}_{z=ai} f(z) + \pi i \mathrm{Res}_{z=-ai} f(z)$$

With this setup we do not actually need to compute $K$ as it must correspond to the imaginary part of the contribution from the two residues. We get for the first residue at $z=ai$ the derivative

$$\frac{1}{z} \frac{1}{(z+ai)^2} \exp(-1/2\times \mathrm{Log}(z)) -2 \mathrm{Log}(z) \frac{1}{(z+ai)^3} \exp(-1/2\times \mathrm{Log}(z)) \\ + \mathrm{Log}(z) \frac{1}{(z+ai)^2} \exp(-1/2\times \mathrm{Log}(z)) \times -\frac{1}{2} \frac{1}{z}.$$

With the branch of the logarithm we find $\mathrm{Log}(ai) = \log a + \pi i/2$, getting

$$\frac{1}{\sqrt{a}} \exp(-\pi i/4) \\ \times \left(\frac{1}{ai} \times - \frac{1}{4 a^2} + (2\log a + \pi i) \frac{1}{8 i a^3} + (\log a + \pi i/2) \times - \frac{1}{4 a^2} \times -\frac{1}{2 ai}\right) \\ = \frac{1}{\sqrt{a}} \exp(-\pi i/4) \frac{1}{8i a^3} (3\log a + 3\pi i/2 - 2).$$

We also have $\mathrm{Log}(-ai) = \log a + 3 \pi i/2$, getting for the second residue at $z=-ai$

$$\frac{1}{\sqrt{a}} \exp(-3\pi i/4) \\ \times \left(- \frac{1}{ai} \times - \frac{1}{4 a^2} - (2\log a + 3\pi i) \frac{1}{8 i a^3} + (\log a + 3\pi i/2) \times - \frac{1}{4 a^2} \times \frac{1}{2 ai}\right) \\ = \frac{1}{\sqrt{a}} \exp(-3\pi i/4) \frac{1}{8i a^3} (- 3\log a - 9\pi i/2 + 2).$$

With $\exp(-\pi i/4) = \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}$ and $\exp(-3\pi i/4) = -\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}$

and factoring out $\frac{1}{\sqrt{a}} \frac{1}{8i a^3}$ we get three contributions, which are

$$\sqrt{2} (3\log a - 2) + 3\pi i (\sqrt{2} + i\sqrt{2}/2)$$

Combine these and multiply by $\pi i$ to get

$$\frac{\pi}{8\sqrt{a} a^3} (\sqrt{2} (3\log a - 2) + 3\pi i (\sqrt{2} + i\sqrt{2}/2)).$$

We extract the real part as promised and obtain

$$\frac{\sqrt{2}\pi}{8\sqrt{a} a^3} (3\log a - 2 - 3\pi/2)$$

or

$$\bbox[5px,border:2px solid #00A000]{ \frac{\pi}{4\sqrt{2} a^{7/2}} \left(3\log a - 2 - \frac{3}{2}\pi\right).}$$

matching the result by @Jack D'Aurizio.

As a bonus we have shown that

$$\int_0^\infty \frac{1}{\sqrt{x} (x^2+a^2)^2} dx = \frac{1}{\pi} \frac{\pi}{8\sqrt{a} a^3} 3\sqrt{2}\pi = \frac{3\pi}{4\sqrt{2} a^{7/2}}.$$

$\endgroup$
1
$\begingroup$

Define $\sqrt{z}$ and $\ln z$ with a branch cut at the positive $x$-axis. Integrate $$f(z) = \frac{\ln z}{\sqrt{z}(z^2+a^2)^2}$$ around keyhole contour. along upper $x$-axis is : $$\int_{\gamma_1} f(z) dz = \int_0^{\infty}\frac{\ln x}{\sqrt{x}(x^2+a^2)^2} dx$$ along lower $x$-axis is: $$\int_{\gamma_2} f(z) dz = \int_\infty^{0}\frac{\ln x+2\pi i}{-\sqrt{x}(x^2+a^2)^2} dx = \int_0^{\infty}\frac{\ln x+2\pi i}{\sqrt{x}(x^2+a^2)^2} dx$$ from which you recovered the original integral.

$\endgroup$
  • $\begingroup$ So using this keyhole contour would mean 2 residues, both of order 2, would be included then? $\endgroup$ – D. Ashton Nov 19 '17 at 11:53
  • 1
    $\begingroup$ You have to calculate the residue of $f(z)$ at $z=ai$ and $z=-ai$, and yes, they're both order $2$. $\endgroup$ – pisco Nov 19 '17 at 11:58
  • $\begingroup$ I get stuck when trying to evaluate $2\pi{i}\int_0^{\infty}\frac{1}{\sqrt{x}(x^2+a^2)^2}dx$. Is there a particular trick to integrating it? $\endgroup$ – D. Ashton Nov 19 '17 at 12:49
1
$\begingroup$

$\phantom{a}$ Dear audience, this is the new episode of Feynman's trick versus contour integration.


Our starting point is $$\forall \alpha\in(-1,1),\qquad \int_{0}^{+\infty}\frac{x^\alpha}{1+x^2}\,dx =\frac{\pi}{2\cos\tfrac{\pi\alpha}{2}}\tag{1}$$ which is a consequence of the substitution $\frac{1}{1+x^2}=u$, Euler's Beta function and the reflection formula for the $\Gamma$ function. We may introduce a further parameter $K>0$ and state $$\forall \alpha\in(-1,1),\forall K>0,\qquad \int_{0}^{+\infty}\frac{x^\alpha}{K+x^2}\,dx =\frac{\pi K^{\frac{\alpha-1}{2}}}{2\cos\tfrac{\pi\alpha}{2}}\tag{2}$$ then differentiate both sides with respect to $K$: $$\forall \alpha\in(-1,1),\forall K>0,\qquad \int_{0}^{+\infty}\frac{x^\alpha}{(K+x^2)^2}\,dx =\frac{\pi(1-\alpha) K^{\frac{\alpha-3}{2}}}{4\cos\tfrac{\pi\alpha}{2}}.\tag{3}$$ Let us differentiate both sides with respect to $\alpha$ (it is more practical to exploit $\frac{df}{dx}=f(x)\cdot\frac{d}{dx}\log f(x)$ to manipulate the RHS): $$\int_{0}^{+\infty}\frac{x^\alpha\log(x)}{(K+x^2)^2}\,dx =\frac{\pi K^{\frac{\alpha-3}{2}}}{8\cos\tfrac{\pi\alpha}{2}}\left[(1-\alpha)\log K+\pi(1-\alpha)\tan\tfrac{\pi\alpha}{2}-2\right].\tag{4}$$ By evaluating $(4)$ at $K=a^2$ and $\alpha=-\frac{1}{2}$ we get: $$\int_{0}^{+\infty}\frac{\log(x)}{\sqrt{x}(x^2+a^2)^2}\,dx =\frac{\pi |a|^{-7/2}}{4\sqrt{2}}\left[3\log|a|-\tfrac{3\pi}{2}-2\right].\tag{5}$$

$\endgroup$
  • $\begingroup$ @FrankW.: in order to reach an integral representation for the Beta function you need to enforce the substitution $x=\sqrt{K} z$, from which the half-integer exponent of $K$ in the closed form. $\endgroup$ – Jack D'Aurizio Jun 8 '18 at 20:25
  • $\begingroup$ ’Aurizio Yeah I realized that soon after I posted my comment. $\endgroup$ – Frank W. Jun 8 '18 at 21:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.