1. ## Itegral

How to integrate
$\int_0^{\infty}{(sechx)^8}dx$

I think it can be solved by $\Gamma .... \beta$ functions
Thanks

2. Originally Posted by Amer
How to integrate
$\int_0^{\infty}{(sech x)^8}dx$

I think it can be solved by $\Gamma .... \beta$ functions
Thanks
let $a > 0$ be a real number and $I_a=\int_0^{\infty} (\text{sech}x)^a \ dx=2^a \int_0^{\infty}(e^x + e^{-x})^{-a} \ dx.$ let $e^{-x}=\tan(t/2).$ then we will have: $I_a=\int_0^{\frac{\pi}{2}} (\sin t)^{a-1} dt=\frac{1}{2}B \left(\frac{a}{2}, \frac{1}{2} \right)=\frac{\sqrt{\pi} \ \Gamma(\frac{a}{2})}{2\Gamma(\frac{a+1}{2})}.$

3. Originally Posted by NonCommAlg
let $a > 0$ be a real number and $I_a=\int_0^{\infty} (\text{sech}x)^a \ dx=2^a \int_0^{\infty}(e^x + e^{-x})^{-a} \ dx.$ let $e^{-x}=\tan(t/2).$ then we will have: $I_a=\int_0^{\frac{\pi}{2}} (\sin t)^{a-1} dt=\frac{1}{2}B \left(\frac{a}{2}, \frac{1}{2} \right)=\frac{\sqrt{\pi} \ \Gamma(\frac{a}{2})}{2\Gamma(\frac{a+1}{2})}.$
After putting $e^{-x} = \tan(t/2)$, how did you get the integrand to be $(\sin t)^{a-1}$?

4. (sin^x/2 + cos^2x/2)(sinx/2 cosx/2)
but
sinx/2 cosx/2 =( sinx)/2 and sin^x/2 + cos^2x/2=1
so

(1/sinx)^-a = (sinx)^a

5. Originally Posted by Amer
(sin^x/2 + cos^2x/2)(sinx/2 cosx/2)
but
sinx/2 cosx/2 =( sinx)/2 and sin^x/2 + cos^2x/2=1
so

(1/sinx)^-a = (sinx)^a

Because you have $2^{a} \int_{0}^{\pi/2} \left(\tan(t/2)+ \frac{1}{\tan(t/2)} \right)^{-a} = 2^{a} \int_{0}^{\pi/2} \left( \frac{2}{\sin t} \right)^{-a} \ dt = \int_{0}^{\pi/2} (\sin t)^{a-1} \ dt$. This follows from identity $\tan \frac{t}{2} = \frac{\sin t}{1+ \cos t}$.

6. Originally Posted by Amer
How to integrate
$\int_0^{\infty}{(sechx)^8}dx$

I think it can be solved by $\Gamma .... \beta$ functions
Thanks
Hi. I haven't carried the integral all the way to the end; however, I believe a simple trig substitution will do. e^2x = tan^2theta

7. Hello, Amer!

Integrate: . $\int_0^{\infty} \text{sech}^8x\,dx$

We have: . $\int\text{sech}^6x\,\left(\text{sech}^2x\,dx\right ) \;=\;\int\left(\text{sech}^2x\right)^3\left(\text{ sech}^2x\,dx\right) \;=\;\int\left(1-\text{tanh}^2x\right)^3\left(\text{sech}^2x\,dx\ri ght)$

. . . . . $= \;\int\left(1 - 3\:\!\text{tanh}^2x + 3\:\!\text{tanh}^4x - \text{tanh}^6x\right)\left(\text{sech}^2x\,dx\righ t)$

Let: $u \:=\:\text{tanh}x \quad\Rightarrow\quad du \:=\:\text{sech}^2x\,dx$

Substitute: . $\int(1 - 3u^2 + 3u^4 - u^6)\,du$ . . . etc.