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It is an interesting exercise to show that the function $f(x)=\sin(x\log x)$ is Riemann-integrable over $\mathbb{R}^+$ (as shown by robjohn in this related question, for instance). Even more interesting is to notice that:

$$\int_{0}^{1}f(x)\,dx = \sum_{n\geq 0}\int_{0}^{1}\frac{(-1)^n x^{2n+1}\left(\log x\right)^{2n+1}}{(2n+1)!}\,dx=\sum_{n\geq 1}\frac{(-1)^n}{(2n)^{2n}},\tag{1}$$ with a series related with the one appearing in many sophomore's dreams.

Moreover, by using Lambert's W function it is not difficult to check that: $$ I = \int_{1}^{+\infty}\sin(x\log x)\,dx = \int_{0}^{+\infty}\frac{\sin u}{1+W(u)}\,du. \tag{2} $$

Now my question: is it possible to give a nice closed form to the RHS of $(2)$ through countour integration, the residue theorem or other techniques?

For instance, is there a closed form for the almost-digamma-sum: $$ g(x)=\sum_{n\geq 0}\left(\frac{1}{1+W(x+2n\pi)}-\frac{1}{1+W(2n\pi)}\right) \tag{3}$$ ? If so, we have just to compute $\int_{0}^{2\pi}g(x)\sin x\,dx$.

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  • $\begingroup$ @Dr.MV: $1+W(x)$ equals zero in $x=-\frac{1}{e}$ and $W(x)$ has a logarithmic growth. Both these facts follow from the definition of Lambert's W function. $\endgroup$ – Jack D'Aurizio Mar 10 '15 at 21:25
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    $\begingroup$ @Jack Do you know this paper: apmaths.uwo.ca/~djeffrey/Offprints/W-adv-cm.pdf there is an excellent discussion of the different branches of $W$ which may help you to find an appropriate contour $\endgroup$ – tired Mar 11 '15 at 14:01
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    $\begingroup$ Wolfy says that $\int_{0.1}^{1000} sin(x\ log(x)) dx = 0.585684$. $\endgroup$ – marty cohen Oct 30 '15 at 21:46
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    $\begingroup$ @martycohen If you replace the upper bound $1000$ by $10^7$ you get $0.4$. Evaluating this integral numerically is very hard to do the slow fall-off and rapid osccilations. $\endgroup$ – Winther Dec 20 '15 at 1:55
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    $\begingroup$ Not Riemann integrable, but such that the improper Riemann integral converges. $\endgroup$ – zhw. Dec 20 '15 at 2:43
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I seem to have an idea for a solution to your integral through other techniques than you might have anticipated. By the way, I am extremely uncertain of my work, so please edit the question before immediately down-voting if you see a problem. Thanks. I want to point out that the final solution is hopefully easier for you. (Point of answer is to solve for $I$)

$$\int_1^\infty\text{d}x \sin(x \log x)=I$$

Now multiply the numerator and denominator of the fraction $I/1$ by J, where $$J=\int_1^e\frac{\text{d}x}{x\log x}$$

From this, you get (with substitution in $J$ of $x=y$) $\frac{IJ}{J}=\cdots$

$$I=\frac{1}{\int_1^e \sqrt{x\log x}}\int_1^\infty \int_1^e \frac{\text{d}x\text{d}x \sin(x\log x)}{\sqrt{x\log x}}$$

Now, the multiplied integral $(J)$ can in fact be solved in terms of the error function. I will solve for $J$ by substitution and Fundamental Theorem of Calculus you will see why later.

$$\int\frac{\text{d}x}{\sqrt{x\log x}}$$

Substitute out $a=\frac{1}{\sqrt{x}}$ and $\text{d}a=-\frac{\text{d}x}{2x^{3/2}}$ I have checked the algebra. $$\int \frac{-2 \text{d}u}{u^2\sqrt{-2\log u}}\implies -i\sqrt{2}\int \frac{\text{d}u}{u^2\sqrt{\log u}}$$

Next substitute $c=\log b$ and $\text{d}c=\frac{\text{d}b}{b}$ That leaves us with

$$-i\sqrt{2}\int \frac{e^{-b}\text{d}b}{\sqrt{b}}$$

The next part is a bit weird. This was beyond my skill level, so I gave this to W|A who gladly accepted. It punched out:

$$-i\sqrt{2\pi} \text{erf}(\sqrt b)$$

Under recollection that $a=\log b$, you get

$$-i\sqrt{2\pi} \text{erf}(\sqrt{\log a})$$

Then recall that $a=\frac{1}{\sqrt{x}}$, you get

$$-i\sqrt{2\pi} \text{erf}(\sqrt{\log \frac{1}{\sqrt{x}}})$$

Simplifying gains:

$$-i\sqrt{2\pi}\text{erf}(\sqrt{\frac{-\log x}{2}})$$

From that, solving for $J$ is a piece of cake.

$$\int_1^e \frac{\text{d}x}{\sqrt{x \log x}}=-i\sqrt{2\pi}(\text{erf}(\sqrt{\frac{-\log e}{2}}))-(-i\sqrt{2\pi}(\text{erf}(\sqrt{\frac{-\log 1}{2}})))=\sqrt{2\pi}\text{erfi}(1/\sqrt{2})\approx 2.38$$

This ugly mess results in a very little change. But in reality, that "very little change" is extremely important. DO NOT change the right side. We have something very special in mind for that.

$$I=\frac{1}{\sqrt {2\pi}\text{erfi}(1/\sqrt{2})}\int_1^\infty \int_1^e \frac{\text{d}x\text{d}x \sin(x\log x)}{\sqrt{x\log x}}$$

So, I do recognize the inverse of $x \log x$ as $W(\log x)$, so for the entire RHS, substitute $x=W(\log u)$

$$I=\frac{1}{\sqrt {2\pi}\text{erfi}(1/\sqrt{2})}\int_{0}^\infty \int_0^{\Omega} \frac{(W^{\prime}(\log u)^2)\text{d}u\text{d}u \sqrt{u}\sin u}{u}$$

We easily recognize something nice! Split up the integral into three parts that can be used. (integral of sinc function is what is recognized.) Not entirely sure of that step (though pretty sure), please let me know in the comments if that is OK to do.

$$I=\frac{1}{\sqrt {2\pi}\text{erfi}(1/\sqrt{2})}*\int_0^\Omega (W^\prime (\log u)^2)\sqrt{u}\text{d}u * \int_0^\infty (\frac{\sin u}{u})\implies$$

$$I=\frac{\sqrt{\pi}}{2\sqrt2 \text{erfi}(1/\sqrt{2})}\int_0^\Omega \sqrt{u}\space W^{\prime}(\ln u)\text{d}u$$

After this step, however, I am at a loss. I hope that this new solution brings you closer to your goal. Sorry for not being able to complete it.

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  • $\begingroup$ @RonGordon Do you mind checking the step my last step? (Seperating Integrals) You are easily the king at integrals and I am trying to learn the properties. $\endgroup$ – user266519 Dec 20 '15 at 1:50
  • $\begingroup$ You have an integral with ${\rm d}x{\rm d}x$. This does not make sense. It should be $\int\int\frac{{\rm d}x\rm{d} y\sin(x\log(x))}{\sqrt{y\log(y)}}$. $\endgroup$ – Winther Dec 20 '15 at 2:00
  • $\begingroup$ The idea behind the $\text{d}x \text{d}x$ is similar to a $\text{d}x \text{d}y$ but with the substitution of $x=y$ But because of the fact that we maintain that fact through out the problem and only substitute the same things for both $x$ and $y$, we are allowed to do that. I hope you can get what I am saying. @Winther $\endgroup$ – user266519 Dec 20 '15 at 2:04
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    $\begingroup$ You are not allowed to do that. $\endgroup$ – Winther Dec 20 '15 at 2:07
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    $\begingroup$ Well, that isn't how I would evaluate $I$ in the first place. I would multiply and divide a helpful integral $J$ that would make it easier. $\endgroup$ – user266519 Dec 20 '15 at 2:12

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