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Definite Integration question

2014 · Shift 2 · Q39
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  5. /2014 · Shift 2 · Q39

Definite Integration question

2014 · Shift 2 · Q39

JEE AdvancedMathematicsDefinite IntegrationMCQ+3 / −1
Given that for each a∈(0,1),   lim⁡h→0+ ∫h1−ht−a(1−t)a−1dta \in \left( {0,1} \right),\,\,\,\mathop {\lim }\limits_{h \to {0^ + }} \,\int\limits_h^{1 - h} {{t^{ - a}}{{\left( {1 - t} \right)}^{a - 1}}dt}a∈(0,1),h→0+lim​h∫1−h​t−a(1−t)a−1dt exists. Let this limit be g(a).g(a).g(a). In addition, it is given that the function g(a)g(a)g(a) is differentiable on (0,1).(0,1).(0,1). The value of g(12)g\left( {{1 \over 2}} \right)g(21​) is
  1. A
    π\piπ
  2. B
    2π2\pi2π
  3. C
    π2{\pi \over 2}2π​
  4. D
    π4{\pi \over 4}4π​
View written solutionFree

Correct answer: A

  1. We are given
g(a)=lim⁡h→0+∫h1−ht−a(1−t)a−1 dt,a∈(0,1).g(a)=\lim_{h\to 0^+}\int_h^{1-h} t^{-a}(1-t)^{a-1}\,dt, \qquad a\in(0,1).g(a)=h→0+lim​∫h1−h​t−a(1−t)a−1dt,a∈(0,1).

We need to find g(12)g\left(\tfrac12\right)g(21​).

  1. First, observe the integrand:
t−a(1−t)a−1=1ta(1−t)1−a.t^{-a}(1-t)^{a-1}=\frac{1}{t^a(1-t)^{1-a}}.t−a(1−t)a−1=ta(1−t)1−a1​.

For a∈(0,1)a\in(0,1)a∈(0,1), near t=0t=0t=0 it behaves like t−at^{-a}t−a, and near t=1t=1t=1 it behaves like (1−t)a−1(1-t)^{a-1}(1−t)a−1. Since

a<1anda−1>−1,a<1 \quad\text{and}\quad a-1>-1,a<1anda−1>−1,

both singularities are integrable. Hence the improper integral converges, so

g(a)=∫01t−a(1−t)a−1 dt.g(a)=\int_0^1 t^{-a}(1-t)^{a-1}\,dt.g(a)=∫01​t−a(1−t)a−1dt.
  1. This is a Beta function integral:
B(x,y)=∫01tx−1(1−t)y−1 dt.B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}\,dt.B(x,y)=∫01​tx−1(1−t)y−1dt.

Match powers:

x−1=−a  ⟹  x=1−a,x-1=-a \implies x=1-a,x−1=−a⟹x=1−a, y−1=a−1  ⟹  y=a.y-1=a-1 \implies y=a.y−1=a−1⟹y=a.

Therefore,

g(a)=B(1−a,a).g(a)=B(1-a,a).g(a)=B(1−a,a).
  1. Use the Beta-Gamma relation:
B(1−a,a)=Γ(1−a)Γ(a)Γ(1).B(1-a,a)=\frac{\Gamma(1-a)\Gamma(a)}{\Gamma(1)}.B(1−a,a)=Γ(1)Γ(1−a)Γ(a)​.

Since Γ(1)=1\Gamma(1)=1Γ(1)=1,

g(a)=Γ(a)Γ(1−a).g(a)=\Gamma(a)\Gamma(1-a).g(a)=Γ(a)Γ(1−a).

Now apply Euler’s reflection formula:

Γ(a)Γ(1−a)=πsin⁡(πa).\Gamma(a)\Gamma(1-a)=\frac{\pi}{\sin(\pi a)}.Γ(a)Γ(1−a)=sin(πa)π​.

So,

g(a)=πsin⁡(πa).g(a)=\frac{\pi}{\sin(\pi a)}.g(a)=sin(πa)π​.
  1. Substitute a=12a=\tfrac12a=21​:
g(12)=πsin⁡(π2)=π1=π.g\left(\frac12\right)=\frac{\pi}{\sin\left(\frac{\pi}{2}\right)}=\frac{\pi}{1}=\pi.g(21​)=sin(2π​)π​=1π​=π.
  1. Hence the correct option is:
π\boxed{\pi}π​

which is Option A.

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