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I'm having trouble to prove the following: $$ \int_{0}^{\infty}\frac{\tan^{a}\left(x\right)} {x^{2} + b^{2}}\,\mathrm{d}x = \frac{\pi}{2b}\,\frac{\tanh^{a}\left(b\right)}{\cos\left(\pi a/2\right)}\quad \mbox{where}\ \left\vert a\right\vert < 1\ \mbox{and}\ b > 0 $$ The main difficulty is that the sign of the $\tan\left(x\right)$ alternates between positive and negative and so we need $\left(-1\right)$ to raise to power of noninteger number $a$.

I have no idea how to proceed.

Felix Marin
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Martin Gales
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    I think the main problem is that $\tan x$ produces infinitely many improperties: $\frac{\pi}{2}+k\pi$ for $k\in\Bbb N$. – Tito Eliatron Dec 08 '20 at 17:23
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    BTW, maybe trying to use Residue Theorem with $f(z)=\frac{\tan z}{z^2+b^2}$ would be an idea – Tito Eliatron Dec 08 '20 at 17:29
  • @TitoEliatron what contour do you recommend – Martin Gales Dec 08 '20 at 17:37
  • $[-R,R]$ plus the upper semicircle centered at the origin and radius $R$. This is classical. What is differentt here is the function, but taking into account that $\Im(\tan(z))=\frac{\sinh(2y)}{\cos(2x)+\cosh(2y)}$, it may works. (or maybe $\tan(z/2)/(z^2+b^2)$). Just guessing... – Tito Eliatron Dec 08 '20 at 17:40
  • No, that contoyur I think will not work since the singularities of $\tan$ would be in the contour. BUT, since those singularities lies in teh positive real line, maybe taking $[-R,0]+[0,Ri]+C$, where $C$ is the quarter of circle. – Tito Eliatron Dec 08 '20 at 17:47
  • Call me nuts, but if there are values for $x$ for which the function does not exist AND these are asymptotic values (non removable discs), how can you just come up with an integral AND expect an answer? – imranfat Dec 08 '20 at 17:53
  • I don't think this formula is provable as it is is. Tan goes negative and the left hand side will have complex contributions for real a. The right hand side stays real. – user321120 Dec 08 '20 at 17:59
  • Besides the jumps at $\left.\vphantom{\Large A} k\pi + \pi/2,\right\vert_{,k\ \in\ \mathbb{N},,,,}$ you have branch-cuts of $\tan^{a}\left(x\right)$ -thanks to the exponent $a$- – Felix Marin Dec 08 '20 at 19:46

3 Answers3

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The integrand has poles at $x=\pm ib$ and singularities on the real axis at $x=(2n+1)\frac{\pi}{2}$ with cuts where the tangent becomes negative. Nevertheless, the integral can be viewed as a complex contour in the upper half-plane with the segment $(-\infty,\infty)$ shifted slightly by $+i0$ to avoid the singularities and make the integrand holomorphic in the upper half-plane except at $x=ib$. Let $$C_R=(-R,R) \cup \gamma_R$$ be the closed contour where $\gamma_R$ is the large semi-circle counter-clockwise in the upper half-plane of radius $R$ s.t. $R\neq (2n+1)\frac{\pi}{2}$. Since the tangent is bounded on $\gamma_R$, the integral over the arc $\gamma_R$ vanishes in the limit $R \rightarrow \infty$. Using the residue theorem for the limiting closed contour $C_\infty$, we arrive at $$\frac{\pi}{b} \, \tan^a(ib) = \int_{C_\infty} \frac{\tan^a(x)}{x^2+b^2} \, {\rm d}x = \int_{-\infty+i0}^{\infty+i0} \frac{\tan^a(x)}{x^2+b^2} \, {\rm d}x = (e^{i\pi a}+1) \int_0^{\infty} \frac{|\tan(x)|^a}{x^2+b^2} \, {\rm d}x \,. \tag{1}$$ Rearranging gives the desired result.

Technically the last step is a bit more intricate than it appears at first sight. When substituting $x\rightarrow -x$ for the segment $(-\infty+i0,i0)$, it becomes $(-i0,\infty-i0)$ which matters because it changes the phase of the tangent. When $\tan(x)$ becomes negative, it has phase $+\pi$ for $\Im(x)>0$ and $-\pi$ for $\Im(x)<0$. This can be seen for e.g. $x\in(-\pi/2+i0,i0)$ and the fact that for $x$ close to $0$, $\tan(x) \approx x$ and hence they share the same phase properties i.e. $$\arg(\tan(x))=\arg(x)=\arg(-1+i0)=\arg(e^{i\pi})=\pi$$ and by the periodicity this translates to $\mathbb{R}$.

Therefore the correct sequence of steps leading to the RHS of (1) is as follows: $$\int_{i0}^{\infty+i0} \frac{\tan^a(x)}{x^2+b^2} \, {\rm d}x + \int_{-i0}^{\infty-i0} \frac{(-\tan(x))^a}{x^2+b^2} \, {\rm d}x \\ =\sum_{n=0}^\infty \int_{n\pi + i0}^{n\pi +\pi/2+i0} \frac{\tan^a(x)}{x^2+b^2} \, {\rm d}x + \sum_{n=0}^\infty \int_{n\pi + \pi/2 + i0}^{n\pi + \pi+i0} \frac{\tan^a(x)}{x^2+b^2} \, {\rm d}x \\ + \sum_{n=0}^\infty \int_{n\pi - i0}^{n\pi +\pi/2 - i0} \frac{(-\tan(x))^a}{x^2+b^2} \, {\rm d}x + \sum_{n=0}^\infty \int_{n\pi + \pi/2 - i0}^{n\pi +\pi - i0} \frac{(-\tan(x))^a}{x^2+b^2} \, {\rm d}x \, .$$ The argument inside $(\dots)^a$ of the first and last term is positive (on this part of the decomposed domain), i.e. the integrand has phase $0$ and both terms can be combined to $\int_0^\infty \frac{|\tan(x)|^a}{x^2+b^2} \, {\rm d}x$. The second term has $\Im(x)>0$ and thus phase $+\pi$. Likewise in the third term $\Im(x)<0$ and hence $-\tan(x)$ has phase $+\pi$ also. Pulling out the phase and using absolute values instead, these two terms combine to $e^{i\pi a} \int_0^\infty \frac{|\tan(x)|^a}{x^2+b^2} \, {\rm d}x$ giving the RHS of (1).

Diger
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if we try letting $x=iy\Rightarrow dx=idy$ and $(0,\infty)\rightarrow(0,\infty)$. We then know that: $$\tan(iy)=i\tanh(y)$$ $$\frac{1}{(iy)^2+b^2}=\frac{1}{b^2-y^2}$$ so our integral becomes: $$\int_0^\infty\frac{i^{a+1}\tanh^a(y)}{b^2-y^2}dy$$ now you can do factor: $b^2-y^2=(b+y)(b-y)$ and also notice that: $$i^{a+1}=(e^{i\pi/2})^{a+1}=\cos\left(\frac{(a+1)\pi}{2}\right)+i\sin\left(\frac{(a+1)\pi}{2}\right)$$

Henry Lee
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This is not an answer. Just an extended comment.

After using partial fractions expansion for cotangent function this integral can be reduced to $$ \int_0^{\pi/2}\frac{\tan^ax}{\cosh 2b-\cos2x}dx=\frac{\pi\tanh^{a}\left(b\right)}{2\cos\left(\pi a/2\right)\sinh 2b}. $$ Then, after the substitution $\tan x=t$, it becomes $$ \int_0^\infty\frac{t^a}{t^2+c^2}dt=\frac{\pi c^{a-1}}{2\cos\left(\pi a/2\right)}, $$ where we have denoted $c=\tanh b$.

Martin Nicholson
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