$\int_0^{\pi/2} \sec^a(t)\,dt= \frac{\sqrt{\pi}}{2\Gamma\left(1-\frac{a}{2}\right)}\Gamma\left(\dfrac{1-a}{2}\right)$

Question

Inside the Wolfram Documentation page for the secant function, an identity is given which involves the gamma function, polygamma function, and Catalan's constant.

Notes on documentation page:

Some special functions can be used to evaluate more complicated definite integrals. For example, polygamma and gamma functions and the Catalan constant are needed to express the following integral: $$\int_0^{\pi/2} \sec^a(t)\,dt= \frac{\sqrt{\pi}}{2\Gamma\left(1-\frac{a}{2}\right)}\Gamma\left(\dfrac{1-a}{2}\right),\quad\text{$\operatorname{Re}(a)<1$} $$

I know that the Gamma function is defined as

$$\Gamma(z) = \int_0^\infty x^{z-1} e^{-x}\, dx, \quad\text{$\operatorname{Re}(z)>0$}$$

and Catalan's constant can be written as

$$G = \beta(2) = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^2} = \frac{1}{1^2} - \frac{1}{3^2} + \frac{1}{5^2} - \frac{1}{7^2} + \frac{1}{9^2} - \cdots$$

but I don't see how this helps. I couldn't find a source in the Wolfram documentation page and couldn't find a duplicate question on Math SE. How did the author of the Wolfram page arrive at this identity?

Answer 1

To derive the identity, it is a straight up application of the Beta function in the form $$\operatorname{B}(m,n) = 2 \int_0^{\frac{\pi}{2}} \cos^{2m - 1} t \sin^{2n - 1} t \, dt.$$

For the secant integral, we have \begin{align} \int_0^{\frac{\pi}{2}} \sec^a t \, dt &= \int_0^{\frac{\pi}{2}} \cos^{-a} t \, dt\\ &= \frac{1}{2} \cdot 2 \int_0^{\frac{\pi}{2}} \cos^{2\left (\frac{1 - a}{2} \right ) - 1} t \sin^{2 \left (\frac{1}{2} \right ) - 1} t \, dt\\ &= \frac{1}{2} \operatorname{B} \left (\frac{1 - a}{2}, \frac{1}{2} \right )\\ &= \frac{1}{2} \frac{\Gamma \left (\frac{1 - a}{2} \right ) \Gamma \left (\frac{1}{2} \right )}{\Gamma \left (\frac{1 - a}{2} + \frac{1}{2} \right )}\\ &= \frac{\sqrt{\pi}}{2 \Gamma \left (1 - \frac{a}{2} \right )} \Gamma \left (\frac{1 - a}{2} \right ), \end{align} for $a < 1$, as required. Here the following well-known results of $$\operatorname{B} (a,b) = \frac{\Gamma (a) \Gamma (b)}{\Gamma (a + b)},$$ together with $\Gamma (\frac{1}{2}) = \sqrt{\pi}$, have been used.

Answer 2

$$I=\int_{0}^{\pi/2} (\cos x)^{-a}~ dx =\frac{1}{2}\frac{\Gamma(\frac{1}{2})~\Gamma(\frac{1-a}{2})}{\Gamma(\frac{2-a}{2})}=\frac{\sqrt{\pi}}{2} \frac{\Gamma(\frac{1-a}{2})}{\Gamma({1-\frac{a}{2}})}.$$ Here we have used HGamma functions and the trigonometric form of the $\beta$-integral.See https://en.wikipedia.org/wiki/Beta_function

Answer 3

Assuming $0 \leq t \leq \frac \pi 2$, the antiderivative is $$\int \sec^a(t)\,dt=\sin (t) \,\, _2F_1\left(\frac{1}{2},\frac{a+1}{2};\frac{3}{2};\sin ^2(t)\right)$$ where appears the gaussian hypergeometric function. $$\int_0^x \sec^a(t)\,dt=\sin (x) \,\, _2F_1\left(\frac{1}{2},\frac{a+1}{2};\frac{3}{2};\sin ^2(x)\right)$$I suppose that they have been considering the asymptotics of $$\sqrt y \,\, _2F_1\left(\frac{1}{2},\frac{a+1}{2};\frac{3}{2};y\right)$$ when $y \to 1^-$.

Have a look at @Semiclassical's answer to this question. Quoting him

If $\Re(c)>\Re(a+b)$, then $$F(a,b;c;1)=\dfrac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}$$

Using the actual values, we end with $$\int_0^{\pi/2} \sec^a(t)\,dt==\frac12\frac{\Gamma \left(\frac{1-a}{2}\right)}{ \Gamma \left(\frac{2-a}{2}\right)}$$