show that $\int_{0}^{\pi/2}\tan^ax \, dx=\frac {\pi}{2\cos(\frac{\pi a}{2})}$

Question

show that $$\int_{0}^{\pi/2}\tan^ax \, dx=\frac {\pi}{2\cos(\frac{\pi a}{2})}$$

I think we can solve it by contour integration but I dont know how.

If someone can solve it by two way using complex and real analysis its better for me.

thanks for all.

Answer 1

Let $u=\tan{x}$, $dx=du/(1+u^2)$. Then the integral is

$$\int_0^{\infty} du \frac{u^a}{1+u^2}$$

This integral may be performed for $a \in (-1,1)$ by residue theory. By considering a contour integral about a keyhole contour about the positive real axis

we find that

$$\left ( 1-e^{i 2 \pi a} \right) \int_0^{\infty} du \frac{u^a}{1+u^2} = i 2 \pi \frac{e^{i \pi a/2}-e^{i 3 a\pi/2}}{2 i}$$

Or

$$\int_0^{\infty} du \frac{u^a}{1+u^2} = \pi \frac{\sin{\pi a/2}}{\sin{\pi a}} $$

From which the sought after result may be found.

ADDENDUM

A little further explanation. Consider the contour integral

$$\oint_C dz \frac{z^a}{1+z^2}$$

where $C$ is the above keyhole contour. This means that the integral may be written as

$$\int_{\epsilon}^R dx \frac{x^a}{1+x^2} + i R \int_0^{2 \pi} d\theta \,e^{i \theta} \frac{R^a e^{i a \theta}}{1+R^2 e^{i 2 \theta}} + \\ e^{i 2 \pi a} \int_R^{\epsilon}dx \frac{x^a}{1+x^2} + i \epsilon \int_0^{2 \pi} d\phi\,e^{i \phi} \frac{\epsilon ^a e^{i a \phi}}{1+\epsilon ^2 e^{i 2 \phi}} $$

We take the limit as $R \to \infty$ and $\epsilon \to 0$ and we recover the expression for the contour integral above.

Answer 2

You can use the beta function

$$ \beta(x,y) = 2\int_0^{\pi/2}(\sin\theta)^{2x-1}(\cos\theta)^{2y-1}\,d\theta=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}, \qquad \mathrm{Re}(x)>0,\ \mathrm{Re}(y)>0 \! $$

to evaluate the integral. In your case

$$ 2x-1=a \implies x=\frac{a+1}{2}\quad 2y-1=-a \implies y=\frac{1-a}{2}. $$

Answer 3

Sorry for being late $$ \begin{aligned} \int_0^{\frac{\pi}{2}} \tan ^a x d x&=\int_0^{\frac{\pi}{2}} \sin ^a x \cos ^{-a} x d x \\ & =\int_0^{\frac{\pi}{2}} \sin ^{2\left(\frac{a+1}{2}\right)-1} x \cos ^{2\left(\frac{-a+1}{2}\right)-1} x d x \\ & =\frac{1}{2} B\left(\frac{a+1}{2}, \frac{-a+1}{2}\right) \\ & =\frac{1}{2} \pi \csc \frac{(a+1) \pi}{2} \\ & \end{aligned} $$ Applying the Euler-reflection property $$ B(x, 1-x)=\pi \csc (\pi x) \quad x \notin \mathbb{Z}, $$ we have $$\boxed{\int_0^{\frac{\pi}{2}} \tan ^a x d x =\frac{\pi}{2 \cos \left(\frac{\pi a}{2}\right)}} $$