Complex limit

Calculus Level 3

lim ⁡ t → ∞ 1 t ∫ 0 ln ⁡ t cos ⁡ ( i x ) d x = a b \large \lim_{t\to\infty} \dfrac1t \int_0^{\ln t} \cos(i x) \, dx = \dfrac ab

If b b is a prime number, find a + b a+b .


Clarification: i = − 1 i=\sqrt{-1} denotes the imaginary unit .


The answer is 3.

Akeel Howell by Akeel Howell , 22, USA

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1 solution

Akeel Howell
Mar 17, 2017

If we work cos ⁡ ( i x ) \cos{(ix)} algebraically, we obtain

cos ⁡ ( i x ) = e i ( i x ) + e − i ( i x ) 2 = e x + e − x 2 = cosh ⁡ ( x ) \cos{(ix)} =\ \large\dfrac{e^{i(ix)}+e^{-i(ix)}}{2} \\ = \dfrac{e^{x}+e^{-x}}{2} = \cosh{(x)}

So the limit becomes lim ⁡ t → ∞ 1 t ∫ 0 ln ⁡ t cosh ⁡ ( x ) d x = lim ⁡ t → ∞ 1 t sinh ⁡ ( x ) ∣ 0 ln ⁡ t = lim ⁡ t → ∞ e ln ⁡ t − e − ln ⁡ t 2 t = lim ⁡ t → ∞ t − 1 / t 2 t = 1 2 ⟹ a + b = 3 \displaystyle\lim_{t \to \infty}{\dfrac{1}{t}\int_{0}^{\ln{t}}{\cosh{(x)}} dx} \\ = \displaystyle\lim_{t \to \infty}{\dfrac{1}{t}\sinh{(x)}\LARGE{|}_{\small{0}}^{\small{\ln{t}}}} = \lim_{t \to \infty}{\dfrac{e^{\ln{t}}-e^{-\ln{t}}}{2t}} \\ = \displaystyle \lim_{t \to \infty}\dfrac{t-1/t}{2t} = \dfrac{1}{2} \implies a+b = \boxed{3}

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