Let a be a positive real number. Evaluate

lim_{n→∞} n (∫_{0}^{a} (\frac{cosh t}{cosh a})^{2n+1} dt)

Answer: 1/(2 tanh a). Proof Let δ_{n} → 0 a sequence to be chosen later.

n (∫_{0}^{a} (\frac{cosh t}{cosh a})^{2n+1} dt) =

= n ∫_{0}^{a−δ_{n}} (\frac{cosh t}{cosh a})^{2n+1} dt + n ∫_{a−δ_{n}}^{a} (\frac{cosh t}{cosh a})^{2n+1} dt.

0 < n ∫_{0}^{a−δ_{n}} (\frac{cosh t}{cosh a})^{2n+1} dt ≤ n(a − δ_{n}) (\frac{cosh(a − δ_{n})}{cosh a})^{2n+1}

since the function cosh x increases for x > 0. Moreover

cosh(a − δ_{n}) = cosh a − δ_{n} sinh a + o(δ_{n})

so

n(a − δ_{n}) (\frac{cosh(a − δ_{n})}{cosh a})^{2n+1} = n(a − δ_{n}) (1 − tanh aδ_{n} + o(δ_{n}))^{2n+1} =

= (a − δ_{n})e^{ln n+(2n+1) ln(1−δ_{n} tanh a+o(δ_{n}))} = = (a − δ_{n})e^{ln n+(2n+1)(−δ_{n} tanh a+o(δ_{n}))} = = (a − δ_{n})e^{ln n+(2n+1)(−δ_{n} tanh a)(1+o(1))} → 0

if we choose δ_{n} in such a way that ln n + (2n + 1)(−δ_{n} tanh a) → −∞ (for instance δ_{n} = an^{− \frac{2}{3}} . We have obtained

lim_{n→∞} n ∫_{0}^{a−δ_{n}} (\frac{cosh t}{cosh a})^{2n+1} dt = 0

Now the integral between a − δ_{n} and a.

n ∫_{a−δ_{n}}^{a} (\frac{cosh t}{cosh a})^{2n+1} dt = n ∫_{a−δ_{n}}^{a} e^{(2n+1)(ln cosh t−ln cosh a)}dt =

= n ∫_{a−δ_{n}}^{a} e^{(2n+1)(ln(cosh a+(t−a) sinh a+O((t−a)^{2}))−ln cosh a)}dt = = n ∫_{a−δ_{n}}^{a} e^{(2n+1)(ln(1+(t−a) tanh a+O((t−a)^{2})))}dt = = n ∫_{a−δ_{n}}^{a} e^{(2n+1)((t−a) tanh a+O((t−a)^{2}))}dt =

The last integral can be bounded as

n ∫_{a−δ_{n}}^{a} e^{(2n+1)((t−a) tanh a+C_{1}(t−a)^{2})}dt ≤

≤ n ∫_{a−δ_{n}}^{a} e^{(2n+1)((t−a) tanh a+O((t−a)^{2}))}dt ≤

≤ n ∫_{a−δ_{n}}^{a} e^{(2n+1)((t−a) tanh a+C_{2}(t−a)^{2})}dt

where C_{1}, C_{2} > 0. Therefore

n ∫_{a−δ_{n}}^{a} e^{(2n+1)((t−a) tanh a}dt ≤

≤ n ∫_{a−δ_{n}}^{a} e^{(2n+1)((t−a) tanh a+O((t−a)^{2})))}dt ≤

≤ n ∫_{a−δ_{n}}^{a} e^{(2n+1)((t−a) tanh a+C_{2}δ_{n}^{2})}dt

Now we assume nδ_{n}^{2} → 0 (δ_{n} = an^{−2/3} is good) and obtain

n ∫_{a−δ_{n}}^{a} e^{(2n+1)((t−a) tanh a}dt ≤

≤ n ∫_{a−δ_{n}}^{a} e^{(2n+1)((t−a) tanh a+O((t−a)^{2})))}dt ≤

≤ n ∫_{a−δ_{n}}^{a} e^{(2n+1)((t−a) tanh a)}dt · e^{(2n+1)C_{2}δ_{n}^{2}} It follows that

lim_{n→∞} n (∫_{0}^{a} (\frac{cosh t}{cosh a})^{2n+1} dt) = lim_{n→∞} n ∫_{a−δ_{n}}^{a} e^{(2n+1)((t−a) tanh a}dt =

= lim_{n→∞} \frac{n(1 − e^{−(2n+1)δ_{n} tanh a})}{(2n + 1) tanh a} = lim_{n→∞} \frac{n}{(2n + 1) tanh a} = \frac{1}{2 tanh a}

since

lim_{n→∞} e^{−(2n+1)δ_{n} tanh a} = 0