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A toughie trigonometric integral

Prove that

Solution

We state two lemmata first.

Lemma 1: Let . It holds that

Proof: A standard proof can be found through Fourier Series. One can expand in Fourier series the function . Another way to prove the identity is to begin from the Weierstrass product , that is . Taking log on both sides we have that

Differentiating we have

Lemma 2: It holds that

Proof: Just differentiate the above identity.

Apply the above lemmata , we have for the initial integral that

1 Comment

1. Another related Fourier series is the following:

Proof: We are using the reflection formula of the digamma function to extract the result. We have successively:

and the result follows.

Of course,