Let $f(z,\eta)$ be an entire function. I need to calculate (numerically) the integral: $$\int\limits _{0}^{\pi}\mbox{Re}\left(\int\limits _{0}^{\pi}f\left(z,\eta\right)d\eta\right)dz$$ Can I switch the inner order and calculate the next integral instead? $$\mbox{Re}\left[\int\limits _{0}^{\pi}\int\limits _{0}^{\pi}f\left(z,\eta\right)d\eta dz\right]$$

  • $\begingroup$ Yes. On $[0,\pi]$, the differentials are real-valued, so you can pull out the $\operatorname{Re}$. $\endgroup$ Jan 1, 2014 at 23:59

1 Answer 1


Yes, due to the linearity of integration.

The justification is as follows: Define $F(z) = \int_0^\pi f(z, \eta) \, d\eta$. Then $F(z) = F_1(z) + i F_2(z)$ where $F_1(z) = \Re F(z)$ and $F_2(z) = \Im F(z)$. On one hand we have

\begin{align} \int_0^\pi \Re \left( \int_0^\pi f(z, \eta) \, d\eta \right) \, dz &= \int_0^\pi \Re F(z) \, dz \\ &= \int_0^\pi F_1(z) \, dz. \end{align}

On the other hand we have

\begin{align} \Re \left( \int_0^\pi \int_0^\pi f(z, \eta) \, d\eta \, dz \right) &= \Re \left( \int_0^\pi F(z) \, dz \right) \\ &= \Re \left( \int_0^\pi (F_1(z) + i F_2(z)) \, dz \right) \\ &= \Re \left( \int_0^\pi F_1(z) \, dz + i \int_0^\pi F_2(z) \, dz \right). \end{align}

Since $F_1(z)$ and $F_2(z)$ are real, and since they are integrated on the interval $[0, \pi] \subset \mathbb{R}$, the two definite integrals are real. Therefore

$$ \Re \left( \int_0^\pi F_1(z) \, dz + i \int_0^\pi F_2(z) \, dz \right) = \int_0^\pi F_1(z) \, dz $$

Therefore the original two expressions are equal, as claimed.

  • $\begingroup$ Awesome! Thanks! you just made me a happier person right now. $\endgroup$
    – Meir
    Jan 3, 2014 at 14:37

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