To determine the value of g ( 1 2 ) g\left(\frac{1}{2}\right) g ( 2 1 ) g ( a ) g(a) g ( a ) g ( a ) = lim h → 0 + ∫ h 1 − h t − a ( 1 − t ) a − 1 d t = ∫ 0 1 t − a ( 1 − t ) a − 1 d t g(a) = \lim_{h \to 0^+} \int_{h}^{1-h} t^{-a} (1 - t)^{a-1} \, dt = \int_{0}^{1} t^{-a} (1 - t)^{a-1} \, dt g ( a ) = lim h → 0 + ∫ h 1 − h t − a ( 1 − t ) a − 1 d t = ∫ 0 1 t − a ( 1 − t ) a − 1 d t
Substituting a = 1 2 a = \frac{1}{2} a = 2 1 g ( 1 2 ) = ∫ 0 1 t − 1 / 2 ( 1 − t ) − 1 / 2 d t = ∫ 0 1 1 t ( 1 − t ) d t g\left(\frac{1}{2}\right) = \int_{0}^{1} t^{-1/2} (1 - t)^{-1/2} \, dt = \int_{0}^{1} \frac{1}{\sqrt{t(1 - t)}} \, dt g ( 2 1 ) = ∫ 0 1 t − 1/2 ( 1 − t ) − 1/2 d t = ∫ 0 1 t ( 1 − t ) 1 d t
We can evaluate this definite integral using the trigonometric substitution:
t = sin 2 θ ⟹ d t = 2 sin θ cos θ d θ t = \sin^2 \theta \implies dt = 2\sin \theta \cos \theta \, d\theta t = sin 2 θ ⟹ d t = 2 sin θ cos θ d θ
For the limits of integration:
When t = 0 t = 0 t = 0 θ = 0 \theta = 0 θ = 0
When t = 1 t = 1 t = 1 θ = π 2 \theta = \frac{\pi}{2} θ = 2 π
Substituting these into the integral:
g ( 1 2 ) = ∫ 0 π / 2 2 sin θ cos θ sin 2 θ ( 1 − sin 2 θ ) d θ g\left(\frac{1}{2}\right) = \int_{0}^{\pi/2} \frac{2\sin \theta \cos \theta}{\sqrt{\sin^2 \theta (1 - \sin^2 \theta)}} \, d\theta g ( 2 1 ) = ∫ 0 π /2 s i n 2 θ ( 1 − s i n 2 θ ) 2 s i n θ c o s θ d θ g ( 1 2 ) = ∫ 0 π / 2 2 sin θ cos θ sin θ cos θ d θ g\left(\frac{1}{2}\right) = \int_{0}^{\pi/2} \frac{2\sin \theta \cos \theta}{\sin \theta \cos \theta} \, d\theta g ( 2 1 ) = ∫ 0 π /2 s i n θ c o s θ 2 s i n θ c o s θ d θ g ( 1 2 ) = 2 ∫ 0 π / 2 1 d θ = 2 ⋅ ( π 2 − 0 ) = π g\left(\frac{1}{2}\right) = 2 \int_{0}^{\pi/2} 1 \, d\theta = 2 \cdot \left( \frac{\pi}{2} - 0 \right) = \pi g ( 2 1 ) = 2 ∫ 0 π /2 1 d θ = 2 ⋅ ( 2 π − 0 ) = π
Thus, the value of g ( 1 2 ) g\left(\frac{1}{2}\right) g ( 2 1 ) π \pi π
Therefore, the correct option is (A) .