The integral ∫₀¹ 1/7^([ 1/x ])dx, where [ . ] denotes the greatest integer function, is equal to

Solution

Maths sir · 4 steps

Cut [0,1] at x = 1/n so [1/x] is constant on each piece, then sum.

1. Split the interval. For 1/(n+1) < x ≤ 1/n we have n ≤ 1/x < n+1, so [1/x] = n. On that piece the integrand is the constant 7⁻ⁿ, and the piece has length 1/n - 1/(n+1).

   I = Σₙ₌₁^∞ 1/7ⁿ(1/n - 1/(n+1))

2. First sum. Use Σₙ₌₁^∞ tⁿ/n = -ln(1-t) with t = 1/7. This gives Σ 1/(n · 7ⁿ) = -ln 6/7 = ln 7/6.
3. Second sum. Write 1/7ⁿ(n+1) = 7 × 1/7ⁿ⁺¹(n+1). The sum over m = n+1 is the full series without its m = 1 term, so it equals 7(ln 7/6 - 1/7) = 7 ln 7/6 - 1.
4. So. I = ln 7/6 - 7 ln 7/6 + 1 = 1 - 6 ln 7/6 = 1 + 6 logₑ 6/7, which is about 0.075. The answer is (a).

The idea to keep. With [1/x] inside an integral, cut the interval at x = 1/n. Each piece gives a constant, and the sum often collapses into the series -ln(1-t).

Answer from the NTA final answer key. Solution by PrepWiser.