Improper convergent integral: differentiation of parameter

Question

I'm interested in the following integral:

$$I = -\int \limits_0^\infty \frac{\sin (\pi x)}{\log (x)} \mathrm{d}x$$

It can be shown by the courtesy of Dirichlet, that this integral converges; there are no singularities, the function $\sin \pi x/\log x$ is in fact continuous, so the only limit that might be the problem is $x \to \infty$, but, from the Dirichlet test, we know that $\sin \pi x$ has a bounded primitive function and $1/\log(x)$ is monotonically decreasing for $x > 1$.

The only thing that've come to my mind as to transform this is to exploit the following property:

$$\frac{\partial}{\partial a} x^a = x^a \log x$$

So I defined the following function:

$$G (a, b) \equiv - \int \limits_0^\infty e^{-b x} x^a \frac{\sin \pi x}{\log x}$$

And now we can see that

$$F (a, b) \equiv \frac{\partial}{\partial a} G (a, b) \equiv - \int \limits_0^\infty e^{-b x} x^a \sin \pi x$$ which is computable and the result is:

$$F (a, b) = -\frac{\Gamma (1+a)}{\left( b^2 + \pi^2 \right)^{(1+a)/2}} \sin \left( (1+a) \arctan \frac{\pi}{b} \right)$$

Now I hope, that the following is true:

$$I = \lim_{b \to 0^+} G(0,b) = \lim_{b \to 0^+} \left[ \int \limits_C^0 F(a, b) \mathrm{d} a \right] = \int \limits_C^0 \lim_{b \to 0^+} F(a, b) \mathrm{d} a $$ $$\lim_{b \to 0^+} F(a, b) = - \frac{\Gamma (1+a)}{\pi^{1+a}} \cos \frac{\pi a}{2}$$

The final step is to find $C$ such that the calculation is easy. The actual value is $C = -1$, as this is Dirichlet integral, yielding $\pi/2$:

$$I = - \frac{\pi}{2} - \int \limits_{-1}^0 \frac{\Gamma (1+a)}{\pi^{1+a}} \cos \frac{\pi a}{2} \mathrm{d} a $$

The result is, however, wrong, because $I > 0$ and the right hand side is negative.

My goal is to express $I$ in terms of a simpler integral (not necessarily analytically) with a better asymptotic behaviour and/or over a finite interval.

Is it possible to compute $I$ efficiently?

Just a side note: it seems to me that it's impossible to practically compute $I$ by the definition, as the $\log(x)$ is increasing crazily slowly - that means the "$\infty$" would have to be very, very big to get a good approximation.

Answer 1

Using the contour

contour integration gives $$ \begin{align} \overbrace{\color{#C00}{\text{PV}\int_0^\infty\frac{e^{i\pi x}}{\log(x)}\,\mathrm{d}x}}^{\substack{\text{integral along the line}\\\text{minus an infinitesimal}\\\text{interval centered at $1$}}}+\overbrace{\vphantom{\int_0^\infty}\ \ \ \ \ \color{#00F}{\pi i}\ \ \ \ \ }^{\substack{\text{integral along}\\\text{an infinitesimal}\\\text{semicircular arc}\\\text{centered at $1$}}} &=\overbrace{\color{#090}{\int_0^\infty\frac{e^{-\pi x}}{\frac\pi2-i\log(x)}\,\mathrm{d}x}}^{\substack{\text{integral along the}\\\text{positive imaginary axis}}}\\ &=\int_0^\infty\frac{e^{-\pi x}\left(\frac\pi2+i\log(x)\right)}{\frac{\pi^2}4+\log(x)^2}\,\mathrm{d}x\tag{1} \end{align} $$ since there are no singularities inside the contour and the integral along the black arc vanishes as the radius tends to $\infty$.

Taking the imaginary part of $(1)$, we get an integral that is far easier to evaluate numerically: $$ \begin{align} \int_0^\infty\frac{\sin(\pi x)}{\log(x)}\,\mathrm{d}x &=-\pi+\int_0^\infty\frac{e^{-\pi x}\log(x)}{\frac{\pi^2}4+\log(x)^2}\,\mathrm{d}x\\[6pt] &\doteq-3.2191900386476588051\tag{2} \end{align} $$

Answer 2

I noticed that the integrand is faily quadratic between integer values, hence

$$-I=\underbrace{\int_0^2\frac{\sin(\pi x)}{\ln(x)}~\mathrm dx}_{\approx-3.64061894952}+\sum_{n=2}^\infty\int_n^{n+1}\frac{\sin(\pi x)}{\ln(x)}~\mathrm dx$$

$$\small\frac{\sin(\pi x)}{\ln(x)}=\frac{(-1)^{n+1}}{t^2\ln^3(t)}\left[t^2\ln^2(t)+t(x-t)\ln(t)+\bigg[\ln(t)+2-\pi^2t^2\ln^2(t)\bigg]\frac{(x-t)^2}2\right]+\mathcal O((x-t)^3),\\t=n+0.5$$

This is a pretty close approximation and gives us the series

$$I\approx3.64061894952+\sum_{n=2}^\infty\underbrace{\frac{(-1)^n}{t^2\ln^3(t)}\left[\left(1-\frac{\pi^2}{24}\right)t^2\ln^2(t)+\frac{\ln(t)}{24}+\frac1{12} \right]}_{a_n}$$

Which converges really slowly since $|a_n|\in\mathcal O(1/\ln(n))$, but by applying an Euler sum, convergence is rapidly sped up:

$$I\approx3.64061894952+\sum_{k=0}^\infty\frac1{2^{k+1}}\sum_{n=0}^k\binom kna_{n+2}$$

---

If one wishes to allow integrals into the sum,

$$I=3.64061894952-\sum_{k=0}^\infty\frac1{2^{k+1}}\sum_{n=0}^k\binom kn\int_{n+2}^{n+3}\frac{\sin(\pi x)}{\ln(x)}~\mathrm dx$$

Evaluating this up to $k=20$, I get

k    I_k
---- ------------
0    3.2884372794
1    3.2399068661
2    3.2264174379
3    3.2219217488
4    3.2202756011
5    3.2196366110
6    3.2193784711
7    3.2192711120
8    3.2192254632
9    3.2192057126
10   3.2191970462
11   3.2191931992
12   3.2191914749
13   3.2191906955
14   3.2191903408
15   3.2191901783
16   3.2191901035
17   3.2191900689
18   3.2191900528
19   3.2191900453
20   3.2191900418

By approximating further out, I get

\begin{align}I&\approx\int_0^{50}\frac{\sin(\pi x)}{\ln(x)}~{\rm d}x+\sum_{n=0}^{19}\frac1{2^{n+1}}\sum_{k=0}^n\binom nk(-1)^k\int_0^1\frac{\sin(\pi t)}{\ln(t+k+50)}~{\rm d}t\\&\approx\underline{-3.219190038646}743\end{align}