Results 1 to 7 of 7

Math Help - Itegral

  1. #1
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093

    Itegral

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

    I think it can be solved by \Gamma .... \beta functions
    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Amer View Post
    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})}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Sampras's Avatar
    Joined
    May 2009
    Posts
    301
    Quote Originally Posted by NonCommAlg View Post
    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} ?
    Last edited by Sampras; May 28th 2009 at 03:47 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    (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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member Sampras's Avatar
    Joined
    May 2009
    Posts
    301
    Quote Originally Posted by Amer View Post
    (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} .
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jan 2009
    Posts
    49
    Quote Originally Posted by Amer View Post
    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
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,914
    Thanks
    778
    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.

    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum