How to prove that $\int_{0}^{\infty}{\sin^4(x)\ln(x)}\cdot{\mathrm dx\over x^2}={\pi\over 4}\cdot(1-\gamma)?$

How to prove that $$\int_{0}^{\infty}{\sin^4(x)\ln(x)}\cdot{\mathrm dx\over x^2}={\pi\over 4}\cdot(1-\gamma).\tag1$$

Here is my attempt:

$$I(a)=\int_{0}^{\infty}{\ln(x)\sin^4(x)\over x^a}\,\mathrm dx\tag2$$

$$I'(a)=\int_{0}^{\infty}{\sin^4(x)\over x^a}\,\mathrm dx\tag3$$

$$I'(2)=\int_{0}^{\infty}{\sin^4(x)\over x^2}\,\mathrm dx\tag4$$

$$I'(2)=\int_{0}^{\infty}{\sin^2(x)\over x^2}\,\mathrm dx-{1\over 4}\int_{0}^{\infty}{\sin^2(2x)\over x^2}\,\mathrm dx=-{\pi\over 2}\tag5$$

Why is this way wrong?

How to prove (1)?

• What is $\gamma$ here? – Rajat Apr 27 '17 at 5:25
• @Rajat Probably the Euler-Mascheroni Constant. – Carl Schildkraut Apr 27 '17 at 5:34
• @Latte' Do I have permission to restore your previous attempt? It's good to have it in the question even if it doesn't work. – Carl Schildkraut Apr 27 '17 at 5:35
• I believe you have evaluated $I'(2)$ correctly, but that doesn't necessarily give $I(2)$. Perhaps you could evaluate $I'(a)$ for all real $a$ and take $I(0)+\int_0^2 I'(a) da$? – Carl Schildkraut Apr 27 '17 at 5:39
• You do the same error here as I pointed out in your previous question. Differentiating gives an extra factor $\ln x$. You divide by $\ln x$ instead of multiply by it. – mickep Apr 27 '17 at 6:30

2 Answers

This is solved essentially in the same way explained in answers to your previous question. As a convenient starting point, I will refer to @Jack D'Aurizio's answer:

$$\int_{0}^{\infty}\frac{1-\cos(kx)}{x^2}\log(x)\,dx = \frac{k\pi}{2}\left(1-\gamma-\log k\right). \tag{1}$$

Now all you have to do is to write

$$\sin^4 x = \frac{1}{2}(1 - \cos(2x)) + \frac{1}{8}(1 - \cos(4x)). \tag{2}$$

I hope that the remaining computation is clear to you.

For your attempt, a correct computation would begin with

$$\frac{d}{da} \int_{0}^{\infty} \frac{\sin^4 x}{x^a} \, dx = - \int_{0}^{\infty} \frac{\sin^4 x}{x^a} \log x \, dx.$$

Notice that you misidentified the derivative of your parametrized integral.

• Oh yes thank you @Sangchul Lee. – gymbvghjkgkjkhgfkl Apr 27 '17 at 6:05
• To be cited by you is a real honour! (+1) – Jack D'Aurizio Apr 27 '17 at 15:00
• Thank you for kind comments! – Sangchul Lee Apr 27 '17 at 15:27

I will outline a self-contained approach, too. By differentiating the integral definition of the $\Gamma$ function, we get the following Lemma: $$\mathcal{L}(\log x) = -\frac{\gamma+\log(s)}{s}\tag{1}$$ and it is not difficult to compute from $(1)$ the Laplace transform of $\sin^4(x)\log(x)$.
By Euler/De Moivre's formula we have $$\sin^4(x) = \frac{1}{16}\left( e^{4ix}+4 e^{2ix}+6+4 e^{-2ix}+e^{-4ix}\right)\tag{2}$$ and by the shift properties of the Laplace transform and $(1)$ we get $$\forall k\in\mathbb{Z},\qquad \mathcal{L}\left(e^{-kix}\log x\right) = -\frac{\gamma+\log(ki+s)}{ki+s}\tag{3}$$ so by $(1),(2),(3)$ and simple algebraic manipulations we get: $$\mathcal{L}\left(\sin^4(x)\log x\right) = -\frac{24\gamma}{64s+20s^3+s^5}+\text{LogTerm}$$ $${\scriptsize\text{LogTerm} = \frac{1}{16} \left(-\frac{6 \log(s)}{s}+\frac{4 \left(4 \arctan\left(\frac{2}{s}\right)+s \log\left(4+s^2\right)\right)}{4+s^2}-\frac{8\arctan\left(\frac{4}{s}\right)+s \log\left(16+s^2\right)}{16+s^2}\right)}\tag{4}$$ and since $\mathcal{L}^{-1}\left(\frac{1}{x^2}\right)=s$, the computation of the original integral boils down to the computation of elementary integrals by integration by parts. For instance, the term $-\frac{\pi\gamma}{4}$ comes from $$\int_0^{+\infty } \frac{24}{(4+s^2)(16+s^2)} \, ds=\frac{\pi}{4}.\tag{5}$$