Beta and Gamma Functions: Questions and Answers

Beta Function:

The beta function

B(m,n) = $\int_0^1$ x^m-1(1-x)^n-1 dx

converges when m,n >0.

Gamma Function:

The gamma function

Γ(n) = $\int_0^\infty$ e^-xx^n-1dx

converges when n >0.

Properties of Beta and Gamma Functions:

The list of the properties of beta and gamma functions is given below:

P1: B(m,n) = B(n,m)

P2: B(m,n) = 2$\int_0^{\pi/2}$ sin^2m-1θ cos^2n-1θ dθ.

P3: B(m,n) = $\int_0^{\infty} \dfrac{x^{m-1}}{(1+x)^{m+n}}dx$ = $\int_0^{\infty} \dfrac{x^{n-1}}{(1+x)^{m+n}}dx$.

P4: B(m,n) = $\int_0^1 \dfrac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}}dx$.

P5: Γ(n+1) = nΓ(n)

P6: Γ(n+1) = n!

P7: B(m,n) = $\dfrac{\Gamma(m) \Gamma(n)}{\Gamma(m+n)}$

P8: Γ(n)Γ(1-n) = cosec mπ, 0<m<1.

Example1:

Show that I = $\int_0^{\pi/2}$ sin^mx cosⁿx dx converges if m>-1 and n>-1.

Answer:

$\int_0^{\pi/2}$ sin^mx cosⁿx dx

= 1/2 × 2 $\int_0^{\pi/2}$ sin^{2(m/2 +1/2)-1}x cos^{2(n/2 +1/2)-1}x dx

= 1/2 B(m/2 +1/2, n/2 +1/2) by P2

So the given integral I converges if m/2 +1/2>0 and n/2 +1/2>0.

⇒ m>-1 and n>-1.

Example2:

Find I = $\int_0^{\pi/2} \sqrt{\tan x} dx$.

Answer:

As tanx=sinx/cosx, we have

I = $\int_0^{\pi/2}$ sin^1/2x cos^-1/2x dx

= 1/2 × 2$\int_0^{\pi/2}$ sin^2×3/4-1x cos^2×1/4-1x dx

= 1/2 B(3/4, 1/4) by P2

= $\dfrac{1}{2} \dfrac{\Gamma(\frac{3}{4}) \Gamma(\frac{1}{4})}{\Gamma(1)}$ by P7

= $\dfrac{1}{2} \Gamma(\frac{1}{4}) \Gamma(1-\frac{1}{4})$

= 1/2 cosec(π/4) by P8

= π/√2.

Example3:

Find the value of gamma 1/2, that is, find Γ(1/2).

Answer:

Putting m=1/2 and n=1/2 in the above property P7, we get that

$B(\frac{1}{2}, \frac{1}{2})=\dfrac{\Gamma(\frac{1}{2}) \Gamma(\frac{1}{2})}{\Gamma(1)}$

⇒ $\Gamma(\frac{1}{2}) = \sqrt{B(\frac{1}{2}, \frac{1}{2})}$ …(*)

Now, by P2, we obtain that

$B(\frac{1}{2}, \frac{1}{2})$ = 2 $\int_0^{\pi/2}$ sin^2×1/2-1θ cos^2×1/2-1θ dθ

= 2 $\int_0^{\pi/2}$ dθ

= π

So from (*), we get that Γ(1/2) =√π.

Example 4:

Show that $\int_1^\infty$ x^-m-1(logx)ⁿ dx converges if m>0 and n>-1.

Answer:

Put x=e^z

∴ dx= e^zdz

x	1	∞
z	0	∞

So $\int_0^\infty$ x^-m-1(logx)ⁿ dx

= $\int_0^\infty$ e^-mz-z zⁿ e^z dz

= $\int_0^\infty$ e^-mz zⁿ dz

= $\int_0^\infty$ e^-mz z^(n+1)-1 dz

= $\dfrac{\Gamma(n+1)}{m^{n+1}}$; it exists when n+1>0 and m>0.

So the given improper integral converges if m>0 and n>-1.

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