#25 Measure Your Calibre

Calculus Level 3

A problem was given by Akeel Howell , Here you get it Complex Limit

Here he says that to find lim t 0 ln t cos ( i x ) d x = a b t \large \lim_{t \to \infty}{\int_{0}^{\ln{t}}{\cos{(ix)}} \, dx} =\ \dfrac{a}{b}t

Now 1st of all we need cos ( i x ) \cos (ix)

So Initially what i did was this

Step 1 cos ( i x ) \cos(ix) = R e ( e i × ( i x ) ) Re(e^{i \times (ix)}) By Euler's Method

Step 2 So I got R e ( e x ) Re(e^{-x})

Step 3 Then as e x e^{-x} is Fully real so R e ( e x ) = e x = cos ( i x ) Re(e^{-x}) = e^{-x} = \cos(ix)

Step 4 But we know cos ( i x ) = c o s h x = e x + e x 2 \cos (ix) = coshx = \dfrac{e^x+e^{-x}}{2}

So whats wrong in evaluating cos ( i x ) \cos (ix) ?

Input your answer as the step number. If you think whole thing is wrong input 10 10 but give reason.


The answer is 1.

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1 solution

Brian Moehring
Mar 20, 2017

While it is true that e i × i x = cos ( i x ) + i sin ( i x ) , e^{i\times ix} = \cos(ix) + i\sin(ix), we should note that sin ( i x ) \sin(ix) is not real, so the real part of this expression is not simply cos ( i x ) \cos(ix) .

Rather, since cos ( i x ) \cos(ix) is real and sin ( i x ) \sin(ix) is purely imaginary [under the assumption x x is real], we see that ( e i × i x ) = cos ( i x ) + i sin ( i x ) . \Re\left(e^{i\times ix}\right) = \cos(ix) + i\sin(ix).

Therefore, Step 1 \boxed{1} has the error.

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