Definite integral with Beta Function

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Definite integral with Beta Function

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Reynan

#1 h Apr 13, 2025, 9:29 PM

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$\int\limits_{0}^{1}\frac{x^{a-\frac{1}{2}}}{(1-x)^a(1+x\tan^2{\theta})^a}dx=\frac{2\Gamma\left(a+\frac{1}{2}\right)\Gamma(1-a)}{\sqrt{\pi}}\cos^{2a}{\theta}\frac{\sin{\left[(2a-1)\theta\right]}}{(2a-1)\sin{\theta}}$ for
$\tan{\theta}<1,-\frac{1}{2}<a<1$ The source should be from Legendre, but I cannot find it.

This post has been edited 1 time. Last edited by Reynan, Apr 14, 2025, 6:20 AM

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#2 h Apr 14, 2025, 4:52 AM

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I'm afraid the formula is not correct. First, at $x>1$ the integrand becomes complex, with the non-zero imaginary part, and should be defined (choosing the branch, etc.). Second, at $a=0$ the integral diverges, though the formula gives a finite value.

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#3 h Apr 14, 2025, 6:10 AM

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I think the integral is supposed to be from $0$ to $1$ .

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Reynan

#4 h Apr 14, 2025, 6:21 AM

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I am sorry it is supposed to be from 0 to 1

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#5 h Apr 18, 2025, 11:15 AM • 2 Y

Y by MS_asdfgzxcvb, Svyatoslav

I believe the correct statement is that for $a\in(-1/2,1)$ and $\theta\in(-\pi/2,\pi/2)$ it's
$\int_0^1\frac{x^{a-\frac12}}{(1-x)^a(1+x\tan^2\theta)^a}\,\mathrm dx = \begin{cases} \frac{2\Gamma\left(a+\frac12\right)\Gamma(1-a)}{\sqrt\pi}\cdot\cos^{2a}\theta\cdot\frac{\sin((2a-1)\theta)}{(2a-1)\sin\theta} & a\ne\frac12,\theta\ne0, \\ 2\theta\cot\theta & a=\frac12,\theta\ne0, \\ \frac{2\Gamma\left(a+\frac12\right)\Gamma(1-a)}{\sqrt\pi} & \theta=0. \end{cases}$
First we prove that for fixed $a$ the LHS is a real-analytic fcn of $\theta$ on $(-\pi/2,\pi/2)$ . For $a\in(-1/2,1)$ , $x\in[0,1]$ and $t\in(0,\infty)$ let
$f(a,x,t) = \frac{x^{a-\frac12}}{(1-x)^a(1+xt)^a}.$ Then it's enough to prove that for fixed $a$ the fcn $\int_0^1f(a,x,t)\,\mathrm dx$ is a real-analytic fcn of $t$ on $(0,\infty)$ . For $n\in\mathbb N$ we have
$\frac{\partial^n f}{\partial^n t}(a,x,t) = \frac{x^{a-\frac12+n}\prod_{k=0}^{n-1}(-a-k)}{(1-x)^a(1+xt)^{a+n}} = n!\binom{-a}n\cdot\frac{x^{a-\frac12+n}}{(1-x)^a(1+xt)^{a+n}},$ and so (because $a-1/2+n$ and $a+n$ are positive)
$\left|\frac{\partial^n f}{\partial^n t}(a,x,t)\right| \le \frac{n!\binom{-a}n}{(1-x)^a}.$ Let $t_0\in(0,\infty)$ . Then by Taylor's theorem there exists $\tau>0$ (depending on $n$ , $a$ , $x$ , $t$ and $t_0$ ) such that
$f(a,x,t)-\sum_{k=0}^n\frac{\partial^k f}{\partial^k t}(a,x,t_0)\cdot\frac{(t-t_0)^k}{k!} = \frac{\partial^{n+1}f}{\partial^{n+1}t}(a,x,\tau)\cdot\frac{(t-t_0)^{n+1}}{(n+1)!},$ and so
$\left|f(a,x,t)-\sum_{k=0}^n\frac{\partial^k f}{\partial^k t}(a,x,t_0)\cdot\frac{(t-t_0)^k}{k!}\right| \le \binom{-a}{n+1}\cdot\frac{(t-t_0)^{n+1}}{(1-x)^a},$ which implies
$\left| \int_0^1f(a,x,t)\,\mathrm dx - \sum_{k=0}^n\left(\int_0^1\frac{\partial^k f}{\partial^k t}(a,x,t_0)\,\mathrm dx\right)\cdot\frac{(t-t_0)^k}{k!} \right|$ $\le \int_0^1\left|f(a,x,t)-\sum_{k=0}^n\frac{\partial^k f}{\partial^k t}(a,x,t_0)\cdot\frac{(t-t_0)^k}{k!}\right|\,\mathrm dx$ $\le \binom{-a}{n+1}(t-t_0)^{n+1}\int_0^1\frac{\mathrm dx}{(1-x)^a} = \frac{\binom{-a}{n+1}(t-t_0)^{n+1}}{1-a}.$ Since $\binom{-a}{n+1}=O(n^{a-1})$ for $n\to\infty$ , if follows that for $|t-t_0|<1$ , $t>0$ it's
$\lim_{n\to\infty} \left| \int_0^1f(a,x,t)\,\mathrm dx - \sum_{k=0}^n\left(\int_0^1\frac{\partial^k f}{\partial^k t}(a,x,t_0)\,\mathrm dx\right)\cdot\frac{(t-t_0)^k}{k!} \right| = 0,$ i.e.
$\int_0^1f(a,x,t)\,\mathrm dx = \sum_{k=0}^\infty\left(\int_0^1\frac{\partial^k f}{\partial^k t}(a,x,t_0)\,\mathrm dx\right)\cdot\frac{(t-t_0)^k}{k!},$ and so $\int_0^1f(a,x,t)\,\mathrm dx$ is indeed real-analytic in $t$ on $(0,\infty)$ (with only a little more effort we can prove that it's in fact analytic on $(-1,\infty)$ , but we don't need that).

Since for fixed $a$ the RHS is also real-analytic in $\theta$ on $(-\pi/2,\pi/2)$ (the singularity at $0$ is removable in both the $a\ne1/2$ and $a=1/2$ cases), it's enough to prove the equation for $\theta\in(-\pi/4,\pi/4)$ .

So let $\theta\in(-\pi/4,\pi/4)$ and also $a\ne1/2$ , $\theta\ne0$ . We have $|x\tan^2\theta|<1$ for $x\in[0,1]$ and so
$\int_0^1\frac{x^{a-\frac12}}{(1-x)^a(1+x\tan^2\theta)^a}\,\mathrm dx$ $= \int_0^1\sum_{n=0}^\infty\binom{-a}n\tan^{2n}\theta\cdot\frac{x^{a-\frac12+n}}{(1-x)^a}\,\mathrm dx$ $= \sum_{n=0}^\infty\binom{-a}n\tan^{2n}\theta\int_0^1\frac{x^{a-\frac12+n}}{(1-x)^a}\,\mathrm dx$ $= \sum_{n=0}^\infty\binom{-a}n\tan^{2n}\theta\cdot B\left(a+\frac12+n,1-a\right)$ $= \sum_{n=0}^\infty\binom{-a}n\tan^{2n}\theta\cdot\frac{\Gamma\left(a+\frac12+n\right)\Gamma(1-a)}{\Gamma\left(\frac32+n\right)}$ $= \sum_{n=0}^\infty\binom{-a}n\tan^{2n}\theta\cdot\frac{\Gamma\left(a+\frac12\right)\Gamma(1-a)\prod_{k=0}^{n-1}\left(a+\frac12+k\right)} {\Gamma\left(\frac32\right)\prod_{k=0}^{n-1}\left(\frac32+k\right)}$ $= \frac{2\Gamma\left(a+\frac12\right)\Gamma(1-a)}{\sqrt\pi} \cdot \sum_{n=0}^\infty\binom{-a}n\tan^{2n}\theta\cdot\frac{\prod_{k=0}^{n-1}\left(a+\frac12+k\right)}{\prod_{k=0}^{n-1}\left(\frac32+k\right)}$ $= \frac{2\Gamma\left(a+\frac12\right)\Gamma(1-a)}{\sqrt\pi} \cdot \sum_{n=0}^\infty(-1)^n\tan^{2n}\theta\cdot\frac{\prod_{k=0}^{n-1}(-2a-2k)(-2a-2k-1)}{2^nn!\prod_{k=0}^{n-1}(2k+3)}$ $= \frac{2\Gamma\left(a+\frac12\right)\Gamma(1-a)}{\sqrt\pi} \cdot \sum_{n=0}^\infty(-1)^n\tan^{2n}\theta\cdot\frac{\prod_{k=0}^{2n-1}(-2a-k)}{(2n+1)!}$ $= \frac{2\Gamma\left(a+\frac12\right)\Gamma(1-a)\cot\theta}{(-2a+1)\sqrt\pi} \cdot \sum_{n=0}^\infty(-1)^n\binom{-2a+1}{2n+1}\tan^{2n+1}\theta$ $= \frac{2\Gamma\left(a+\frac12\right)\Gamma(1-a)\cot\theta}{2i(-2a+1)\sqrt\pi} \cdot \left[\sum_{n=0}^\infty\binom{-2a+1}{n}(i\tan\theta)^n-\sum_{n=0}^\infty\binom{-2a+1}{n}(-i\tan\theta)^n\right]$ $= \frac{2\Gamma\left(a+\frac12\right)\Gamma(1-a)\cot\theta}{2i(-2a+1)\sqrt\pi} \cdot \left[(1+i\tan\theta)^{-2a+1}-(1-i\tan\theta)^{-2a+1}\right]$ $= \frac{2\Gamma\left(a+\frac12\right)\Gamma(1-a)\cot\theta}{2i(-2a+1)\sqrt\pi} \cdot \left[(\cos^{-1}\theta\cdot e^{\theta i})^{-2a+1}-(\cos^{-1}\theta\cdot e^{-\theta i})^{-2a+1}\right]$ $= \frac{2\Gamma\left(a+\frac12\right)\Gamma(1-a)\cot\theta}{2i(-2a+1)\sqrt\pi} \cdot \left[\cos^{2a-1}\theta\cdot e^{(-2a+1)\theta i}-\cos^{2a-1}\theta\cdot e^{-(-2a+1)\theta i}\right]$ $= \frac{2\Gamma\left(a+\frac12\right)\Gamma(1-a)\cot\theta\cos^{2a-1}\theta\cdot2i\sin((-2a+1)\theta)}{2i(-2a+1)\sqrt\pi}$ $= \frac{2\Gamma\left(a+\frac12\right)\Gamma(1-a)}{\sqrt\pi}\cdot\cos^{2a}\theta\cdot\frac{\sin((2a-1)\theta)}{(2a-1)\sin\theta}.$
The other two cases then follow either by taking limits or can be easily computed directly (the $a=1/2$ case by the third Euler substitution and the $\theta=0$ case is just the Beta fcn).

Remark: the condition $\theta\in(-\pi/4,\pi/4)$ implies $|\tan\theta|<1$ and this is the case that we proved directly, but there are no issues at $\theta=\pm\pi/4$ in neither side of the equation and so (unless I'm completely blundering something) the result indeed extends to $\theta\in(-\pi/2,\pi/2)$ .

This post has been edited 7 times. Last edited by GreenKeeper, Apr 18, 2025, 2:58 PM

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