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We were told today by our teacher (I suppose to scare us) that in certain schools for physics in Soviet Russia there was as an entry examination the following integral given

$$\int\limits_{0}^{\infty} \frac{x^{-\mathfrak{i}a}}{x^2+bx+1} \,\mathrm{d}x\,,$$

where $a \in \mathbb{R}$, $b \in [0,2)$, and $\mathfrak{i}$ is the imaginary unit. And since we are doing complex analysis at the moment, it can, according to my teacher, be calculated using complex methods.

I was wondering how this could work? It seems hard to me to find a good curve to apply the residue theorem for this object, I suppose. Is there a trick to compute this integral?

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    $\begingroup$ A keyhole contour. On $\mathbb{C}\setminus [0,+\infty)$, write $z^{-ia} = \exp(-ia \log z)$. Then $\lvert z^{-ia}\rvert \leqslant \exp (2\pi\lvert a\rvert)$. $\endgroup$ – Daniel Fischer Nov 20 '17 at 19:14
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    $\begingroup$ @ThePirateBay: Mathematica has to be trained to consider the right constraints and perform the correct simplifications. The answer is elementary, but it is not that easy to find. $\endgroup$ – Jack D'Aurizio Nov 20 '17 at 19:21
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    $\begingroup$ @MathematicianByMistake: I totally agree. With some experience in converting series into integrals (and viceversa) via $\mathcal{L}/\mathcal{L}^{-1}$ such integral is a rather innocent-looking one, but do they teach the Mellin transform in Russian high schools? $\endgroup$ – Jack D'Aurizio Nov 20 '17 at 19:36
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    $\begingroup$ @MathematicianByMistake: In Soviet Russia, integral tackles you! $\endgroup$ – Alex R. Nov 21 '17 at 0:22
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    $\begingroup$ Yes I showed this question to my mother. She chuckled and a tear of nostalgia came to her eye. Thanks for this. $\endgroup$ – Sentinel Nov 21 '17 at 0:37
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$\phantom{}$ Dear MSE users, this is the new episode of Mister Feynman and Monsieur Laplace versus contour integration.

Tonight we have a scary integral, but we may immediately notice that $$ \mathcal{L}(x^{-ia})(s) = s^{ia-1}\Gamma(1-ia),\qquad \mathcal{L}^{-1}\left(\frac{1}{x^2+bx+1}\right)(s) =\frac{e^{Bs}-e^{\overline{B}s}}{\sqrt{b^2-4}}$$ where $B$ is the root of $x^2+bx+1$ with a positive imaginary part. By the properties of the Laplace transform, the original integral is converted into $$\frac{\Gamma(1-ia)}{\sqrt{b^2-4}}\int_{0}^{+\infty}s^{ia-1}\left(e^{Bs}-e^{\overline{B}s}\right)\,ds$$ which can be evaluated in terms of the $\Gamma$ function.
Due to the reflection formula, the final outcome simplifies into $$ \frac{\left(B^{-i a}-\overline{B}^{-i a}\right) \pi }{\sinh(\pi a)\sqrt{4-b^2}} $$ and we may notice that $B=\exp\left[i\arccos\frac{b}{2}\right]$ allows a further simplification.

Outro: poor children.

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    $\begingroup$ This is a truly impressive answer. Are you sure that you're human? $\endgroup$ – BobaFret Nov 21 '17 at 4:35
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    $\begingroup$ @BobaFret Do you have evidence than any nonhumans could produce an answer of this quality? $\endgroup$ – jpmc26 Nov 21 '17 at 5:25
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    $\begingroup$ +1 for a great opening line. And the rest of the answer is remarkable. $\endgroup$ – Paramanand Singh Nov 21 '17 at 8:36
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    $\begingroup$ This is exactly the approach I take with most daunting integrals, in the hopes it one day works. Seems one does get lucky sometimes :) $\endgroup$ – JamalS Nov 21 '17 at 14:49
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    $\begingroup$ @ThePirateBay: just done :) $\endgroup$ – Jack D'Aurizio Nov 21 '17 at 16:58
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$\newcommand{\Res}{\text{Res}}$ Firstly define: \begin{align} f(z) = \frac{1}{z^2+bz+1} \end{align} Secondly define: \begin{align} g(z)= (-z)^{-ia}f(z) \end{align} We use the Principal Log to define $(-z)^{-ia}$. The main reason for the minus sign is that I want to work with the principal Log. Consider now the keyhole contour $K_R$ consisting of $C_R \cup C_R^+ \cup C_R^-$. The circle part is $C_R$ with radius $R$, and $C_R^+$ is the segment that connects $0$ to $R$ on the right half plane from above and $C_R^-$ the same as $C_R^+$ but from below.

So our contour looks like the one below:

enter image description here

Note that $\int_{C_R} g(z)dz \to 0$ as $R\to \infty$ (why?). Now the poles of this function $g(z)$ is at $z_1= -\frac{1}{2}b - i\ \sqrt[]{1-\frac{1}{4}b^2}$ and $z_2 = -\frac{1}{2}b + i\ \sqrt[]{1-\frac{1}{4}b^2}$. Note that for large enough $R$ both poles will be in the area enclosed by our contour.

On $C_R^+$ we have: \begin{align} \lim_{R\to \infty} \int_{C_R^+} g(z) dz = e^{-a\pi}\int^{\infty}_0\frac{x^{-ia}}{x^2+bx+1} dx \end{align} Similarly on $C_R^-$ we have: \begin{align} \lim_{R\to \infty} \int_{C_R^-} g(z) dz = -e^{a\pi}\int^{\infty}_0\frac{x^{-ia}}{x^2+bx+1} dx \end{align} So: \begin{align} \lim_{R\to\infty} \int_{K_R} g(z) dz &= (-e^{a\pi} + e^{-a\pi} )\int^\infty_0 \frac{x^{-ia}}{x^2+bx+1} dx \\ &=-2\sinh(a\pi) \int^\infty_0 \frac{x^{-ia}}{x^2+bx+1} dx \end{align} On the other hand we have by the Residue theorem: \begin{align} \lim_{R\to\infty} \int_{K_R} g(z) dz = 2\pi i\left( \Res_{z=z_1}g(z)+\Res_{z=z_2}g(z)\right) \end{align} Let's calculate the residues: \begin{align} \Res_{z=z_1} g(z) = \frac{\left(\frac{1}{2}b + i\ \sqrt[]{1-\frac{1}{4}b^2}\ \right)^{-ia}}{-2i \ \sqrt[]{1-\frac{1}{4}b^2} } \end{align} The other one: \begin{align} \Res_{z=z_2} g(z) = \frac{\left(\frac{1}{2}b - i\ \sqrt[]{1-\frac{1}{4}b^2}\ \right)^{-ia}}{2i \ \sqrt[]{1-\frac{1}{4}b^2} } \end{align} Define $\beta:=\frac{1}{2}b + i\ \sqrt[]{1-\frac{1}{4}b^2}$. So: \begin{align} \lim_{R\to\infty} \int_{K_R} g(z) dz = \pi \frac{\bar\beta^{-ia}-\beta^{-ia}}{\sqrt[]{1-\frac{1}{4}b^2}} = 2\pi \frac{\bar\beta^{-ia}-\beta^{-ia}}{ \sqrt[]{4-b^2}} \end{align} This means: \begin{align} \int^{\infty}_0 \frac{x^{-ia}}{x^2+bx+1}dx &= \frac{2\pi}{-2\sinh(a\pi)} \frac{\bar\beta^{-ia}-\beta^{-ia}}{ \sqrt[]{4-b^2}} \end{align} Note that $\beta = \exp(i\arccos(b/2))$. So $\beta^{-ia}=\exp(a\arccos(b/2))$ and $\bar\beta^{-ia}=\exp(a\arccos(b/2)).$ Substituting $\beta$ gives us the final result:

\begin{align} \int^{\infty}_0 \frac{x^{-ia}}{x^2+bx+1}dx = \color{red}{\frac{2\pi\sinh(a\cdot\arccos(b/2))}{\sinh(a\pi) \ \sqrt[]{4-b^2}} } \end{align}

This integral is real.


Remark

It may be interesting to note the following. For positive $x$ one has: $x^{-ia}=e^{-ia\text{Log}(x)}=e^{-ia\ln(x)}=\cos(a\ln(x))-i\sin(a\ln(x))$. So we have: \begin{align} \int^\infty_0 \frac{x^{-ia}}{x^2+bx+1} dx = \int^\infty_0 \frac{\cos(a\ln(x))}{x^2+bx+1} dx - i\int^\infty_0 \frac{\sin(a\ln(x))}{x^2+bx+1} dx \end{align} Since our integral is real we get the following identities for free: \begin{align} \color{blue}{\int^\infty_0 \frac{\cos(a\ln(x))}{x^2+bx+1} dx = \frac{2\pi\sinh(a\cdot\arccos(b/2))}{\sinh(a\pi) \ \sqrt[]{4-b^2}}} \end{align} And: \begin{align} \color{blue}{\int^\infty_0 \frac{\sin(a\ln(x))}{x^2+bx+1} dx = 0 } \end{align}

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    $\begingroup$ The last integral is $0$ and it can be proved by splitting the integral over $[0,1]$ and $[1,\infty)$ and putting $t=1/x$ in second integral. $\endgroup$ – Paramanand Singh Nov 21 '17 at 8:39
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    $\begingroup$ I'd assume this was the expected answer. $\endgroup$ – nbubis Nov 21 '17 at 10:57
  • $\begingroup$ @nbubis yeah it seems that I went for the answer on "I was wondering how this (complex method) could work?" - OP and Jack for the answer on "Is there a trick to compute this integral?" - OP. What a fantastic man! $\endgroup$ – Shashi Nov 21 '17 at 11:27
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    $\begingroup$ Although I love Jack's answer, this answer is exactly what the OP was asking for. I'm "voting" for this one! $\endgroup$ – Mark Fischler Nov 24 '17 at 16:28
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In a complex analysis class, you’re more likely to come across the integral $$\int_{0}^{\infty} \frac{x^{\alpha}}{1+2x \cos \beta +x^{2}} \, dx = \frac{\pi \sin (\alpha \beta)}{\sin(\alpha \pi) \sin(\beta)}, \quad (-1<\alpha <1, \ 0 < \beta < \pi).$$

See this answer, for example, which uses a semicircle in the upper half-plane that is indented at the origin.

Alternatively, you could also use the keyhole contour that Shashi used.

Also see this question about the peculiar symmetry of this integral when it's written in a slightly different form.

The function on the right is not defined at $\alpha =0$. But you could show separately (by using the same contour or by completing the square) that the value of the integral at $\alpha =0$ is $$\lim_{a \to 0} \frac{\pi \sin (\alpha \beta)}{\sin(\alpha \pi) \sin(\beta)} = \frac{\beta}{\sin \beta}.$$

Now if we assign the function on the right the value $\frac{\beta}{\sin \beta}$ at $\alpha =0$, then right side of the equation is a holomorphic function for $-1 <\operatorname{Re}(\alpha) <1$ with $\beta$ fixed.

And since the integral on the left is absolutely convergent in the strip $-1 < \operatorname{Re}(\alpha) < 1$, we can use a property of the Mellin transform that states that the integral defines an holomorphic function in that strip.

(This is very similar to property of the Laplace transform mentioned here, and can be proved in essentially the same manner.)

So by the identity theorem, the formula holds for $-1 <\operatorname{Re}(\alpha) <1$.

Your integral is the case $\alpha=-ia$ and $\cos(\beta) = \frac{b}{2}$.

If $b\in [0, 2)$, then $\beta$ falls between means $0$ and $\pi$, and we get

$$\begin{align} \int_{0}^{\infty} \frac{x^{-ia}}{x^{2}+bx+1} \, dx &= \frac{\pi \sin\left(-ia \arccos\left(\frac{b}{2} \right)\right)}{\sin(-ia \pi) \sin \left(\arccos \left(\frac{b}{2} \right) \right)} \\ &= \frac{\pi \sinh \left(a \arccos\left(\frac{b}{2} \right) \right)}{\sinh(a \pi)\frac{\sqrt{4-b^{2}}}{2}}. \end{align} $$

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