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I saw these two intgerals and would like to know if they are correct.

$$\int_{0}^{\infty}{\sin(e^{-\gamma}x)}\cdot{\ln{x}\over x}\mathrm dx=0\tag1$$

$$\int_{0}^{\infty}{\sin\left(\sqrt{x}^{\sqrt{2}}\right)}\cdot{\ln{x}\over x}\mathrm dx=-\pi\gamma\tag2$$

This is the well know $\gamma$ Euler-Mascheroni's constant

Here I ignored the limits

I apply sub: to $(1)$

$u=\ln{x}$

$xdu=dx$

$$I = \int{ue^{-u}\sin(e^{u-\gamma})}du\tag3$$

Apply integration by parts to $(3)$

$$\int{ue^{-u}}du=-e^{-u}(1+u)$$

$$I={-e^{-u}(1+u)\sin(e^{u-\gamma})}+\int{e^{-u}(1+u)\cos(e^{u-\gamma})}du$$

Encounter more harder than before.

Please show us how to prove $(1)$ and $(2)$

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  • $\begingroup$ $u = \ln x \implies du = \frac 1x dx$ so your new integral should just be $$\int u \sin(e^{u-\gamma}) du$$ $\endgroup$ – mattos Jan 17 '17 at 12:37
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First note that

$$\int_{0}^{\infty}{\sin(x)}\cdot{\ln{x}\over x}\mathrm dx=-\frac{\pi \gamma}{2}$$

Note the Mellin transform of the sine function

$$\int^\infty_0 x^{s-1}\sin(x)\,dx = \Gamma (s) \sin\left( \frac{\pi s}{2} \right)$$

Which can be written as

$$\int^\infty_0 x^{s-1}\sin(x)\,dx = \frac{\pi \Gamma(s+1)}{2\Gamma(s/2+1)\Gamma(1-s/2)}$$

Now differentiate with respect to $s$ and let $s \to 0$ to obtain the result.

First integral

\begin{align} \int_{0}^{\infty}{\sin(ax)}\cdot{\ln{x}\over x}\mathrm dx&=\int_{0}^{\infty}{\sin(t)}\cdot{\ln{t}-\ln a\over t}\mathrm dt\\&= \int_{0}^{\infty}{\sin(t)}\cdot{\ln{t}\over t}\mathrm dt-(\ln a) \frac{\pi}{2} \\ &=\frac{-\pi \gamma}{2}-\ln(a) \frac{\pi}{2} \end{align}

Now letting $a =e^{-\gamma}$ the result follows.

Second Integral

$$\int_{0}^{\infty}{\sin\left(x^{a}\right)}\cdot{\ln{x}\over x}\mathrm dx =\frac{1}{a^2}\int_{0}^{\infty}{\sin\left(t\right)}\cdot{\ln{t}\over t}\mathrm dt = -\frac{\gamma \pi}{2a^2}$$

For $a = \sqrt{2}/2$ the result follows.

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