calculus - How to prove that $int_{0}^{infty}{sin^4(x)ln(x)}cdot{mathrm dxover x^2}={piover 4}cdot(1-gamma)?$

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)?

Answer

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.

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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.