Possible result of integral $\int_{0}^{\infty}\left(x^{\frac{\sin\left(\pi x\right)}{x}}-1\right)dx=0$: Too good to be true? This question is far beyond my level but let me propose it for the aesthetic. First the result have been found with Desmos so there is a possibility to have a wrong result.
Claim:
$$\int_{0}^{\infty}\left(x^{\frac{\sin\left(\pi x\right)}{x}}-1\right)dx=^?0$$
It doesn't seems to be trivial at first glance but make me a comment if so.
Some tought :
Following an idea due to user Momo here Prove or disprove that : $x^{\frac{\sin\left(x\right)}{x}}> \sin\left(x\right)+\frac{1}{x-1}$ for $x\geq \pi$
We have using the well-know identity on the real $e^x\geq x+1$ :
$$x^\frac{\sin(x)}{x}\geq 1+\ln(x)\frac{\sin(x)}{x}$$
To have a better  approximation we integrate on $[1,\infty)$ it gives :
$$\int_{0}^{\infty}\left(x^{\frac{\sin\left(\pi x\right)}{x}}-1\right)dx\geq I=\int_{0}^{1}\left(x^{\frac{\sin\left(\pi x\right)}{x}}-1\right)dx+\int_{1}^{\infty}1+\ln(x)\frac{\sin(x)}{x}-1 dx$$
It gives us :
$$I\simeq 0.075122-0.616781=-0.541659$$
Now using the result due to user Sangchul lee we see that the integral converges but it's not equal to zero (so be careful with software like Desmos in a such case)
Question: Can we hope to find a closed form or an interesting series like the Sophomore's dream ?
 A: This will be a “long comment”. I tried a few techniques, but none were the right way. It seems like the main way to complete it is through basic algebra and integral properties. It is harder than it looks.
Integral of function
Using odd and even rules of functions brings back the original integral. Note that sin(-x)=sin(x), hence it is an even function.
$$\mathrm{f(x)=Even(x)+Odd(x)=\frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}\implies I=\int_0^\infty x^\frac{sin(\pi x)}{x}-1dx=\frac12\int_0^\infty  x^\frac{sin(\pi x)}{x}+ (-x)^\frac{sin(\pi x)}{x}+x^\frac{sin(\pi x)}{x}-(-x)^\frac{sin(\pi x)}{x}-2dx}$$
There is a way to put it into another form using the Abel-Plana formula. Note the integrand value has the limit value of -1 at x=0. The sum part of the formula has all index>0 terms as 0, so it is written here already:
$$\mathrm{\int_0^\infty x^\frac{sin(\pi x)}{x}-1dx=-\frac32+i\int_0^\infty \frac{(ix)^\frac{sinh(\pi x)}{x}-(-ix)^\frac{sinh(\pi x)}{x}}{e^{2\pi x}-1}dx\mathop =^?0}$$
$\large{\text{Possible Proof:}}$
$$\mathrm{\int_0^\infty x^\frac{sin(\pi x)}{x}-1dx=\lim_{b,n\to\infty,b\gg n}\frac bn \sum_{k=1}^n \left[\left(\frac{kb}{n}\right)^{\frac{sin \left(\frac{\pi kb}{n}\right)}{\frac{kb}{n}}} -1\right] =0}$$
This uses the Riemann Sum definition of an integral. Assuming b is a large number or $\infty$, b can always be made larger than n. In the limit, the graph gets closer to 0 with larger n. Even though the sum had a greater chance of evaluating to 0 for larger n, there still is a chance for an n chosen to evaluate to a partial sum S with $|S|\le N$ with N as the upper bound for S. This chance for evaluation to another number other than 0 decreases as $n\to \infty$. Therefore, when b and n both approach infinity with the upper integral bound b growing arbitrarily larger than, the partial sum S gets closer to 0. This means that the integral is 0.

Please correct me and give me feedback!
A: Not an answer, but a strong indication that the integral is not zero.
Evaluating such integrals numerically with Mathematica is a bit tricky, because a rather obvious call to NIntegrate[..., {x,0,Infinity}] usually leads nowhere. Instead, it is often better to evaluate the integrals over sufficiently many finite intervals and see how the integral changes.
So, here is the initial code:
ClearAll[int];
int[0] = 0;
For[k = 1, k <= 100, k++,
  int[k] = int[k-1] +
    NIntegrate[
      Exp[Log[x] Sin[Pi x] / x] - 1,
      {x, 1000 (k-1), 1000 k}
    ]
]
ListPlot[int /@ Range[100]]

It stores the integrals over $[0, 1000k]$ as int[k] and then plots the results for $k$ equal to $1$ through $100$.
However, Mathematica complains about insufficient speed of convergence, and suggests increasing recursion limit and working precision. Let us do so:
ClearAll[int];
int[0] = 0;
For[k = 1, k <= 100, k++,
  int[k] = int[k-1] +
    NIntegrate[
      Exp[Log[x] Sin[Pi x] / x] - 1,
      {x, 1000 (k-1), 1000 k},
      MaxRecursion -> 30,
      WorkingPrecision -> 30
    ]
]
ListPlot[int /@ Range[100]]

Now this takes a bit longer to evaluate (nearly five minutes on my laptop), but luckily Mathematica's complains are gone. Here is what comes out:

Interestingly, with the above method, the result does not seem to depend essentially on the fine-tuning parameters: we have
int[100] = -0.160888472353482801858094440731

with the above choice, and
int[100] = -0.160889

with default parameters (and lots of warning messages). For the record:
NIntegrate[Exp[Log[x] Sin[Pi x]/x] - 1, {x, 0, Infinity}] = -40.5999

is completely off, and
NIntegrate[Exp[Log[x] Sin[Pi x]/x] - 1, {x, 0, Infinity},
  MaxRecursion -> 30, WorkingPrecision -> 30] = 34.8902020565215540843737276445

is not any better. One can likely play with other integration strategies and get a better result, but I did not try to do that.
