I have the following integral: I = ∫ 0 1 d x cos ⁡ ( e 2 π i x ) .

Question

I have the following integral:  I  =      ∫          0              1        d  x    cos  ⁡      (          e              2        π        i        x              )    .

Harold Cantrell · Accepted Answer

After the change of variable   x  →  u defined by   u  =      e          2            π            i            x       you arrive to      ∫    0    1    cos  ⁡  (      e          2            π            i            x        )  d  x  =      1          2            π            i            ∫    C              cos      ⁡      (      u      )        u    d  uwhere C es the unit circle. Now, applying Residue theorem to above integral      1          2            π            i            ∫    C              cos      ⁡      (      u      )        u    d  u  =      1          2            π            i        ∗  2    π    i    ∗  r  e  s  (            cos      ⁡      (      u      )        u    ,  0  )  =      1          2            π            i        ∗  2    π    i  ∗  1  =  1

Sonia Gay · Accepted Answer

You need to use contour integration for complex numbers. If you have a function

f : C \to C,

then one has that

\int_{γ} f (z) d z = \int_{t_{0}}^{t_{1}} f (γ (t)) γ^{'} (t) d t,

where

γ : [t_{0}, t_{1}] \to C

is a differentiable function on

(t_{0}, t_{1}) .

Here, what you have is

\int_{0}^{1} \cos (e^{2 π x i}) d x = \int_{0}^{1} \frac{\cos (e^{2 π x i})}{2 π i e^{2 π x i}} 2 π i e^{2 π x i} d x = \frac{1}{2 π i} \int_{0}^{1} \frac{\cos (e^{2 π x i})}{e^{2 π x i}} 2 π i e^{2 π x i} d x .

The situation here suggests that we have

γ (x) = e^{2 π x i}

and

f (z) = \frac{\cos (z)}{z} .

With that, you have

\frac{1}{2 π i} \int_{0}^{1} \frac{\cos (e^{2 π x i})}{e^{2 π x i}} 2 π i e^{2 π x i} d x = \frac{1}{2 π i} \int_{γ} \frac{\cos (z)}{z} d z .

Notice that the contour

γ

is a closed curve in the complex plane, with

γ (0) = γ (1),

and it has winding number 1. Therefore, by Cauchy's residue theorem, one has that

\frac{1}{2 π i} \int_{γ} \frac{\cos (z)}{z} d z = R e s (0, \frac{\cos (z)}{z}) = lim_{z \to 0} z \frac{\cos (z)}{z} = lim_{z \to 0} \cos (z) = \cos (0) = 1.

The reason the substitution you used does not work is because you failed to account for the fact that your substitution is a closed curve enclosing a singularity at 0.

I have the following integral: I = <msubsup> ∫ <mrow class="MJX-TeX

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