Beta and Gamma Functions: Questions and Answers

Beta Function:

The beta function

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

converges when m,n >0.

Gamma Function:

The gamma function

Γ(n) = $\int_0^\infty$ e-xxn-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$\int_0^{\pi/2}$ sin2m-1θ cos2n-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.


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


$\int_0^{\pi/2}$ sinmx cosnx dx

= 1/2 × 2 $\int_0^{\pi/2}$ sin2(m/2 +1/2)-1x cos2(n/2 +1/2)-1x 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.


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


As tanx=sinx/cosx, we have

I = $\int_0^{\pi/2}$ sin1/2x cos-1/2x dx

= 1/2 × 2$\int_0^{\pi/2}$ sin2×3/4-1x cos2×1/4-1x 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.


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


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 $\int_0^{\pi/2}$ sin2×1/2-1θ cos2×1/2-1θ 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.


Put x=ez

∴ dx= ezdz


So $\int_0^\infty$ x-m-1(logx)n dx

= $\int_0^\infty$ e-mz-z zn ez 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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