$\int_{[0,2\pi]}f d\lambda=\int_{[\alpha,\alpha+2\pi]}f d\lambda$ for complex $f$ with $f(x)=f(x+2\pi)$

Let $$f$$ be a complex-valued, measurable function defined on $$\mathbb R$$ with $$f(x)=f(x+2\pi)$$

I want to show $$\int_{[0,2\pi]}f d\lambda=\int_{[\alpha,\alpha+2\pi]}f d\lambda$$

$$(\alpha \in \mathbb R)$$

My original attempt was to write $$\int_{[\alpha,\alpha+2\pi]}f d\lambda -\int_{[0,2\pi]}f d\lambda=\int_{\alpha}^{\alpha+2\pi}f(x)dx-\int_{0}^{2\pi}f(x)dx=F(\alpha+2\pi)-F(2\pi)-(F(\alpha)-F(0))=\int_{2\pi}^{\alpha+2\pi}f(x)dx-\int_{0}^{\alpha}f(x)dx=0$$

But I think I can't simply write is a Riemann Integral.

How can I show $$\int_{[0,2\pi]}f d\lambda=\int_{[\alpha,\alpha+2\pi]}f d\lambda$$ instead?

• Simply split $[\alpha,\alpha+2\pi] = [\alpha,2\pi] \cup [2\pi , 2\pi+\alpha]$. – Yanko Jan 12 at 16:04
• How does that help? – user626880 Jan 12 at 16:12
• I post a more detailed answer. – Yanko Jan 12 at 16:16
• – user587192 Jan 12 at 16:44

Edit: This proof assumes that $$\alpha<2\pi$$. But since $$f$$ is $$2\pi$$ periodic the function $$f$$ on $$[\alpha,2\pi+\alpha]$$ behaves the same as $$f$$ on $$[\alpha-2\pi,\alpha]$$ and we can keep removing $$2\pi$$ until $$\alpha-2m\pi<2\pi$$, then denote $$\beta=\alpha-2m\pi$$ and argue as below with $$\beta$$ instead of $$\alpha$$:

We can write $$[\alpha,\alpha+2\pi] = [\alpha,2\pi]\cup [2\pi,2\pi+\alpha]$$. Since the union is disjoint (up to one element) we have

$$\int_\alpha^{\alpha+2\pi} f(x)dx = \int_\alpha^{2\pi} f(x)dx+\int_{2\pi}^{2\pi+\alpha} f(x)dx$$

By assumption $$f(x)=f(x+2\pi)$$ and so by changing variables the second integral becomes

$$\int_{2\pi}^{2\pi+\alpha} f(x)dx = \int_0^\alpha f(x+2\pi)dx = \int_0^\alpha f(x)dx$$

Insert to the first equality we have

$$\int_\alpha^{\alpha+2\pi} f(x)dx = \int_\alpha^{2\pi} f(x)dx+\int_{0}^{\alpha} f(x)dx = \int_0^{2\pi}f(x)dx$$

• What if $\alpha>2\pi$? How do you make sense of the interval $[\alpha,2\pi]$? – user587192 Jan 12 at 16:37
• This would work if, instead of your "$2\pi$", you used the (unique) integer multiple of $2\pi$ between $\alpha$ and $\alpha+2\pi$. – paul garrett Jan 12 at 16:40
• @user587192 If $\alpha>2\pi$ then move the entire interval by $2\pi$ as many times as needed. – Yanko Jan 12 at 16:54