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

∫10f(x)dx=∑n≥0∫10(−1)nx2n+1(logx)2n+1(2n+1)!dx=∑n≥1(−1)n(2n)2n,

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=∫+∞1sin(xlogx)dx=∫+∞0sinu1+W(u)du.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)=∑n≥0(11+W(x+2nπ)−11+W(2nπ))

? If so, we have just to compute ∫2π0g(x)sinxdx.

**Answer**

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)

∫∞1dxsin(xlogx)=I

Now multiply the numerator and denominator of the fraction I/1 by J, where J=∫e1dxxlogx

From this, you get (with substitution in J of x=y) IJJ=⋯

I=1∫e1√xlogx∫∞1∫e1dxdxsin(xlogx)√xlogx

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.

∫dx√xlogx

Substitute out a=1√x and da=−dx2x3/2

I have checked the algebra.

∫−2duu2√−2logu⟹−i√2∫duu2√logu

Next substitute c=logb and dc=dbb

That leaves us with

−i√2∫e−bdb√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√2πerf(√b)

Under recollection that a=logb, you get

−i√2πerf(√loga)

Then recall that a=1√x, you get

−i√2πerf(√log1√x)

Simplifying gains:

−i√2πerf(√−logx2)

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

∫e1dx√xlogx=−i√2π(erf(√−loge2))−(−i√2π(erf(√−log12)))=√2πerfi(1/√2)≈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=1√2πerfi(1/√2)∫∞1∫e1dxdxsin(xlogx)√xlogx

So, I do recognize the inverse of xlogx as W(logx), so for the entire RHS, substitute x=W(logu)

I=1√2πerfi(1/√2)∫∞0∫Ω0(W′(logu)2)dudu√usinuu

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=1√2πerfi(1/√2)∗∫Ω0(W′(logu)2)√udu∗∫∞0(sinuu)⟹

I=√π2√2erfi(1/√2)∫Ω0√u W′(lnu)du

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.

**Attribution***Source : Link , Question Author : Jack D’Aurizio , Answer Author : Community*