Fungsi Gamma


Definisi.

Fungsi Gamma \Gamma, suatu fungsi bernilai riil dengan satu peubah, didefinisikan oleh suatu bentuk integral, yaitu :

\Gamma (n) = \displaystyle \int_0^{\infty} x^{n-1} e^{-x} ~dx, n > 0

Berikut beberapa sifat fungsi gamma :

  1. \Gamma (1) = 1

    BUKTI.

    \Gamma (1) = \displaystyle \int_0^{\infty} x^{1-1} e^{-x} ~dx

    = \displaystyle \lim_{M \to \infty} \int_0^{M} e^{-x} ~dx

    = \lim_{M \to \infty} \left( -e^{-x} \right) \mid_{x=0}^M

    = \lim_{M \to \infty} \left( -\dfrac{1}{e^M} + \dfrac{1}{e^0} \right)

    = 0 + 1

    = 1

  2. \Gamma \left( \dfrac{1}{2} \right) = \sqrt{\pi}

    BUKTI.

    \Gamma \left( \dfrac{1}{2} \right) = \displaystyle \int_0^{\infty} x^{\frac{1}{2}-1} e^{-x} ~dx

    = \displaystyle \int_0^{\infty} x^{-\frac{1}{2}} e^{-x} ~dx

    gunakan integral substitusi, misal x = u^2 maka dx = 2u ~du. Sehingga diperoleh x^{-\frac{1}{2}} = (u^2)^{-\frac{1}{2}} = u^{-1}.

    batas-batas integral : x = 0 \rightarrow u = 0

    x = \infty \rightarrow u = \infty

    = \displaystyle \int_0^{\infty} u^{-1} e^{-u^2} 2u ~du

    = \displaystyle 2 \int_0^{\infty} e^{-u^2} ~du

    = \displaystyle 2 \int_0^{\infty} e^{-v^2} ~dv

    \left( \Gamma \left( \dfrac{1}{2} \right) \right)^2 = \left(2 \displaystyle \int_0^{\infty} e^{-u^2} ~du \right) \left( 2 \int_0^{\infty} e^{-v^2} ~dv \right)

    = \displaystyle 4 \int_0^{\infty} \int_0^{\infty} e^{-(u^2+v^2)} ~du ~dv

    transformasi ke koordinat kutub

     photo fungsigamma_zps0cee7830.jpg

    ambil : u = r \cos \theta dan v = r \sin \theta sehingga berakibat u^2 + v^2 = r^2. Oleh karena itu, didapat

    = 4 \displaystyle \int_0^{\frac{\pi}{2}} \int_0^{\infty} e^{-r^2} \begin{vmatrix} \frac{\partial u}{dr} & \frac{\partial v}{dr}\\ \frac{\partial u}{d \theta} & \frac{\partial v}{d \theta} \end{vmatrix} ~dr ~d \theta

    = 4 \displaystyle \int_0^{\frac{\pi}{2}} \int_0^{\infty} e^{-r^2} \begin{vmatrix} \cos \theta & -r \sin \theta\\ \sin \theta & r \cos \theta \end{vmatrix} ~dr ~d \theta

    = 4 \displaystyle \int_0^{\frac{\pi}{2}} \int_0^{\infty} e^{-r^2} r ~dr ~d\theta

    gunakan kembali integral substitusi, ambil A = r^2 maka dA = 2r ~dr. Sehingga diperoleh,

    = \displaystyle 4 \int_0^{\frac{\pi}{2}} \int_{r=0}^{r=\infty} e^{-A} ~\dfrac{1}{2} ~dA ~d \theta

    = \displaystyle 4 \int_0^{\frac{\pi}{2}} \left( -\dfrac{1}{2} e^{-A} \right) \mid_{r=0}^{r=\infty} ~d \theta

    = \displaystyle 4 \int_0^{\frac{\pi}{2}} \dfrac{1}{2} \left[ -\dfrac{1}{e^{\infty}} + \dfrac{1}{e^0} \right] ~d \theta

    = \displaystyle 4 \int_0^{\frac{\pi}{2}} \dfrac{1}{2} ~d \theta

    = \displaystyle 2 \int_0^{\frac{\pi}{2}} ~d \theta

    = 2 \dfrac{\pi}{2}

    = \pi

  3. \Gamma (n+1) = n \Gamma (n)

    BUKTI.

    \Gamma (n+1) = \displaystyle \int_0^{\infty} x^{(n+1)-1} ~e^{-x} ~dx

    = \displaystyle \int_0^{\infty} x^n ~e^{-x} ~dx

    dengan menggunakan integral parsial, ambil u = x^n \Rightarrow du = n x^{n-1} ~dx dan dv = e^{-x} ~dx \Rightarrow v = -e^{-x}

    = \displaystyle uv -\int_0^{\infty} v ~du

    = \displaystyle -x^n e^{-x} -\int_0^{\infty} n x^{n-1} (-e^{-x}) ~dx

    = \displaystyle \lim_{M \to \infty} \dfrac{-x^n}{e^x} \mid_{x=0}^{M} + n \int_0^{\infty} x^{n-1} e^{-x} ~dx

    INGAT definisi : \Gamma (n) = \int_0^{\infty} x^{n-1} e^{-x} ~dx

    = \lim_{M \to \infty} \left[ \dfrac{-M^n}{e^M} + \dfrac{-0^n}{e^0} \right] + n \Gamma (n)

    = \lim_{M \to \infty} \left[ \dfrac{-M^n}{e^M} \right] + n \Gamma (n)

    gunakan dalil L’Holpotal untuk \lim_{M \to \infty} \left[ \dfrac{-M^n}{e^M} \right], maka

    = \lim_{M \to \infty} \left[ \dfrac{-n.M^{n-1}}{e^M} \right] + n \Gamma (n)

    = \lim_{M \to \infty} \left[ \dfrac{-n(n-1)M^{n-2}}{e^M} \right] + n \Gamma (n)

    jika diturunin terus menerus, maka diperoleh

    = \lim_{M \to \infty} \left[ \dfrac{-n!}{e^M} \right] + n \Gamma (n)

    = 0 + n \Gamma (n)

    = n \Gamma (n)

  4. \Gamma (n+1) = n!

    BUKTI.

    \Gamma (n+1) = n \Gamma (n) [berdasarakan sifat ke-3]

    = n \Gamma ((n-1)+1)

    = n(n -1) \Gamma (n-1)

    = n(n -1) \Gamma ((n-2)+1)

    = n(n -1)(n -2) \Gamma (n-2)

    jika dilakukan terus menerus, diperoleh

    = n(n -1)(n -2) \ldots 2 \Gamma (1)

    = n(n -1)(n -2) \ldots (2)(1) [berdasarkan sifat 1]

    = n!

10 comments on “Fungsi Gamma

  1. Ping-balik: Luas Kurva Distribusi Normal | Math IS Beautiful

  2. Assalamualaikumm… trima kasih kak banyak bermanfaat .. kalau boleh ooreksi sdikit .. fungsi gamma stengahnya mungkin hasilnya akar phhi… krn sudah td dikuadratkankan nilai gammanya sblumnya.. trima kasih sbelumnya..

  3. Assalamualaikum.. kak.. trima kasih banyak kak.. sgt bermanfaat…

    Dan mungkin nilai gamma stengahnya itU akar phi.. krn sblumnya sdah dikuadratjkan.. trimakasih .. slainnya semua ktrenfansgt jelas penjelasannya

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