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Let $f: [a,b] \to \mathbb C, t \to f(t) = \text {Re } f(t) + i\text{ Im } f(t)$. Suppose $f$ is continuous.

Let $\int\limits_a^b f(t)dt = \rho e^{i\theta}$, where $\rho \geq 0$ is the module. Then: $$|\int\limits_a^b f(t)dt| = \rho = e^{-i\theta}\int\limits_a^b f(t)dt = \text{Re }\int\limits_a^b e^{-i\theta}f(t)dt = \int\limits_a^b \text{Re }(e^{-i\theta}f)(t)dt \leq \int\limits_a^b |f(t)|dt$$

I don't understand 2 steps:

  1. $e^{-i\theta}\int\limits_a^b f(t)dt = \text{Re }\int\limits_a^b e^{-i\theta}f(t)dt$

  2. $\int\limits_a^b \text{Re }(e^{-i\theta}f)(t)dt \leq \int\limits_a^b |f(t)|dt$

user2016
  • 21
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2 Answers2

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The first step isn't right, and it should have $+i\theta$ rather than $-i\theta$. This is because

$$\rho = e^{i\theta} \big(\rho e^{-i\theta}\big) = e^{i\theta} \int_a^b f(t) \, dt$$

Now take real parts - since $\rho$ is real, the right hand side is real too.

The second step is just an application of the fact that the real part is bounded by the modulus, together with the fact that $|e^{i\theta}| = 1$.

1

For the first step, up to the sign, note that $\rho$ is real. Since it is equal to $e^{-i\theta}\int\limits_a^b f(t)dt$, the last expression is a real number and so $$\rho=e^{-i\theta}\int\limits_a^b f(t)dt= \text{Re }e^{-i\theta}\int\limits_a^b f(t)dt.$$

For the second step, write $z=x-iy$ and note that $$\text{Re }z=x\le |x|\le \sqrt{x^2}\le\sqrt{x^2+y^2}=|z|.$$

John B
  • 16,854