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I can't calculate the following integral

$$\int_{0}^{\infty}\ln\Big(\frac{\sin^2(x)}{x^2}+1\Big)dx=?$$

I can prove that it converges because:

$$\forall x\geq 0\quad \ln(x+1)\leq x$$

So : $$\int_{0}^{\infty}\ln\Big(\frac{\sin^2(x)}{x^2}+1\Big)dx<\int_{0}^{\infty} \frac{\sin^2(x)}{x^2}dx=\frac{\pi}{2}$$

Logically proceeding from my bound, I tried using power series without any success. See Wolfram alpha for more details. I think that integration by parts can give something interesting, but I cannot go further with that. I think it's not a hard integral but I cannot solve it.

Any help is greatly appreciated.

Thanks in advance for all your contributions.

Anay Mehrotra
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Miss and Mister cassoulet char
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1 Answers1

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Note that $$\ln(1+t) = \sum_{n = 1}^\infty (-1)^{n-1} \frac{t^n}{n}$$ so \begin{align*} I &= \int_0^\infty \ln \left( \frac{\sin^2 x}{x^2} +1 \right) dx \\ & = \int_0^\infty \sum_{n = 1}^\infty (-1)^{n-1} \frac{1}{n}\frac{\sin^{2n} x}{x^{2n}}dx = \sum_{n = 1}^\infty (-1)^{n-1} \frac{1}{n}\int_0^\infty\frac{\sin^{2n} x}{x^{2n}}dx \end{align*} One can refer this post to have \begin{align*} \int_0^\infty \frac{\sin^{2n}x}{x^{2n}}dx &= \frac{\pi}{2^{2n} (2n-1)!} \sum_{k=0}^{n} (-1)^k {2n \choose k} (2n-2k)^{2n-1} \\ & = \frac{n \pi }{ (2n)!} \sum_{k=0}^{n} (-1)^k {2n \choose k} (n-k)^{2n-1} \end{align*} So \begin{align*} I & = \pi \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{ (2n)!} \sum_{k=0}^{n} (-1)^k {2n \choose k} (n-k)^{2n-1} \end{align*}

I cannot simplify this more. When I can do I will edit this post.

dust05
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