1. Soal pertama meminta kita menggunakan hubungan rekursi fungsi gamma untuk menyederhanakan \(\Gamma\left(\frac{7}{2}\right)\) dan \(\Gamma\left(-\frac{7}{2}\right)\).
2. Hubungan rekursi fungsi gamma adalah:
$$\Gamma(z+1) = z \Gamma(z)$$
Ini berarti kita bisa menurunkan argumen fungsi gamma dengan mengalikan dengan faktor yang sesuai.
3. Untuk \(\Gamma\left(\frac{7}{2}\right)\):
\begin{align*}
\Gamma\left(\frac{7}{2}\right) &= \left(\frac{5}{2}\right) \Gamma\left(\frac{5}{2}\right) \\
&= \left(\frac{5}{2}\right) \left(\frac{3}{2}\right) \Gamma\left(\frac{3}{2}\right) \\
&= \left(\frac{5}{2}\right) \left(\frac{3}{2}\right) \left(\frac{1}{2}\right) \Gamma\left(\frac{1}{2}\right)
\end{align*}
4. Kita tahu \(\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}\), jadi:
$$\Gamma\left(\frac{7}{2}\right) = \frac{5}{2} \times \frac{3}{2} \times \frac{1}{2} \times \sqrt{\pi} = \frac{15}{8} \sqrt{\pi}$$
5. Untuk \(\Gamma\left(-\frac{7}{2}\right)\), gunakan hubungan:
$$\Gamma(z) = \frac{\Gamma(z+1)}{z}$$
Jadi:
\begin{align*}
\Gamma\left(-\frac{7}{2}\right) &= \frac{\Gamma\left(-\frac{5}{2}\right)}{-\frac{7}{2}} = \frac{\Gamma\left(-\frac{3}{2}\right)}{-\frac{7}{2} \times -\frac{5}{2}} = \frac{\Gamma\left(-\frac{1}{2}\right)}{-\frac{7}{2} \times -\frac{5}{2} \times -\frac{3}{2}} \\
&= \frac{\Gamma\left(\frac{1}{2}\right)}{-\frac{7}{2} \times -\frac{5}{2} \times -\frac{3}{2} \times -\frac{1}{2}}
\end{align*}
6. Menghitung penyebut:
$$-\frac{7}{2} \times -\frac{5}{2} = \frac{35}{4}, \quad \frac{35}{4} \times -\frac{3}{2} = -\frac{105}{8}, \quad -\frac{105}{8} \times -\frac{1}{2} = \frac{105}{16}$$
7. Jadi:
$$\Gamma\left(-\frac{7}{2}\right) = \frac{\sqrt{\pi}}{\frac{105}{16}} = \frac{16}{105} \sqrt{\pi}$$
8. Soal kedua meminta menyatakan integral sebagai fungsi gamma dan menghitung numeriknya.
9. Integral umum:
$$\int_0^\infty x^{a} e^{-x} dx = \Gamma(a+1)$$
10. Untuk (a):
$$a = \frac{2}{3} \implies \int_0^\infty x^{\frac{2}{3}} e^{-x} dx = \Gamma\left(\frac{2}{3} + 1\right) = \Gamma\left(\frac{5}{3}\right)$$
11. Untuk (b):
$$a = -\frac{2}{5} \implies \int_0^\infty x^{-\frac{2}{5}} e^{-x} dx = \Gamma\left(-\frac{2}{5} + 1\right) = \Gamma\left(\frac{3}{5}\right)$$
12. Menghitung nilai numerik dengan komputer (hasil aproksimasi):
$$\Gamma\left(\frac{5}{3}\right) \approx 0.9027, \quad \Gamma\left(\frac{3}{5}\right) \approx 1.4892$$
13. Soal ketiga menggunakan fungsi beta dan hubungannya dengan gamma.
14. Definisi fungsi beta:
$$B(x,y) = \int_0^1 t^{x-1} (1-t)^{y-1} dt = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}$$
15. Untuk (a):
$$I = \int_0^\infty \frac{x^3}{(1+x)^6} dx$$
16. Substitusi: \(t = \frac{x}{1+x} \Rightarrow x = \frac{t}{1-t}\), dan diferensial:
$$dx = \frac{dt}{(1-t)^2}$$
17. Integral menjadi:
\begin{align*}
I &= \int_0^1 \left(\frac{t}{1-t}\right)^3 (1 + \frac{t}{1-t})^{-6} \frac{dt}{(1-t)^2} \\
&= \int_0^1 t^3 (1-t)^{-3} (1-t)^6 \frac{dt}{(1-t)^2} = \int_0^1 t^3 (1-t)^{6-3-2} dt = \int_0^1 t^3 (1-t)^1 dt
\end{align*}
18. Ini adalah fungsi beta dengan:
$$x = 4, \quad y = 2$$
19. Jadi:
$$I = B(4,2) = \frac{\Gamma(4) \Gamma(2)}{\Gamma(6)}$$
20. Menghitung gamma:
$$\Gamma(4) = 3! = 6, \quad \Gamma(2) = 1! = 1, \quad \Gamma(6) = 5! = 120$$
21. Maka:
$$I = \frac{6 \times 1}{120} = \frac{1}{20} = 0.05$$
22. Untuk (b):
$$I = \int_0^1 \frac{x^4}{\sqrt{1-x^2}} dx = \int_0^1 x^4 (1-x^2)^{-\frac{1}{2}} dx$$
23. Substitusi: \(t = x^2 \Rightarrow dt = 2x dx \Rightarrow dx = \frac{dt}{2x}\)
24. Integral menjadi:
\begin{align*}
I &= \int_0^1 x^4 (1 - t)^{-\frac{1}{2}} \frac{dt}{2x} = \frac{1}{2} \int_0^1 x^3 (1 - t)^{-\frac{1}{2}} dt \\
&= \frac{1}{2} \int_0^1 t^{\frac{3}{2}} (1 - t)^{-\frac{1}{2}} dt
\end{align*}
25. Ini adalah fungsi beta dengan:
$$x = \frac{5}{2}, \quad y = \frac{1}{2}$$
26. Jadi:
$$I = \frac{1}{2} B\left(\frac{5}{2}, \frac{1}{2}\right) = \frac{1}{2} \frac{\Gamma\left(\frac{5}{2}\right) \Gamma\left(\frac{1}{2}\right)}{\Gamma(3)}$$
27. Menghitung gamma:
$$\Gamma\left(\frac{5}{2}\right) = \frac{3}{2} \times \frac{1}{2} \times \sqrt{\pi} = \frac{3}{4} \sqrt{\pi}, \quad \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}, \quad \Gamma(3) = 2! = 2$$
28. Jadi:
$$I = \frac{1}{2} \times \frac{\frac{3}{4} \sqrt{\pi} \times \sqrt{\pi}}{2} = \frac{1}{2} \times \frac{3}{4} \times \frac{\pi}{2} = \frac{3 \pi}{16}$$
Jawaban akhir:
\begin{align*}
&1a) \quad \Gamma\left(\frac{7}{2}\right) = \frac{15}{8} \sqrt{\pi} \\
&1b) \quad \Gamma\left(-\frac{7}{2}\right) = \frac{16}{105} \sqrt{\pi} \\
&2a) \quad \int_0^\infty x^{\frac{2}{3}} e^{-x} dx = \Gamma\left(\frac{5}{3}\right) \approx 0.9027 \\
&2b) \quad \int_0^\infty x^{-\frac{2}{5}} e^{-x} dx = \Gamma\left(\frac{3}{5}\right) \approx 1.4892 \\
&3a) \quad \int_0^\infty \frac{x^3}{(1+x)^6} dx = \frac{1}{20} = 0.05 \\
&3b) \quad \int_0^1 \frac{x^4}{\sqrt{1-x^2}} dx = \frac{3 \pi}{16}
\end{align*}
Gamma Beta Integrals Ad07F4
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