# Challenging integral $\int_{0}^{1}\frac{x\operatorname{li}(x)}{x^2+1}dx$

Find the value of the following integral :

$$I=\int_{0}^{1}\frac{x\operatorname{li}(x)}{x^2+1}dx=?\tag{1}$$

where $$\operatorname{li}(x)$$ is the Logarithmic Integral Function

I have tried to use integration by parts :

$$\int_{0}^{1}\frac{x\operatorname{li}(x)}{x^2+1}dx=[\frac{1}{2}\operatorname{li}(x)\ln(x ^2+1)]_0^1-\int_{0}^{1}\frac{\operatorname{ln}(x^2+1)}{\ln(x)}dx\tag{2}$$

But it doesn't converge.

So the second idea is to use power series we have :

$$\frac{1}{x^2+1}=1-x^2+x^4-x^6+x^8-x^{10}+x^{12}+\cdots$$

Moreover we have :

$$\int_{0}^{1}\operatorname{li}(x)x^ndx=-\frac{\ln(n+2)}{n+1}\tag{3}$$

One proof of this is given here by @Zacky

So we get an alternating series that I cannot evaluate .

$$I=\frac{1}{2} \sum _{n=1}^{\infty } \frac{(-1)^{n} \log (2 n+1)}{n}\tag{4}$$

Any help is greatly appreciated .

## Some experimentation

The integral is equivalent to :

$$\int_{0}^{\frac{\pi}{4}}\tan(x)\operatorname{li}(\tan(x))dx\tag{5}$$

Differentiating under the integral the expression:

$$\tan(x)\operatorname{li}(\tan(x))$$

Becomes :

$$\frac{1}{\cos^2(x)}\operatorname{li}(\tan(x))+\frac{1}{\cos^2(x)}\frac{\tan(x)}{\log(\tan(x))}$$

Performing the substitution $$x=\arctan(t)$$

We get (under the integral) :

$$\operatorname{li}(x)+\frac{x}{\log(x)}$$

I don't know what to do next (even if it was funny as experimentation).Maybe I make forbidden things ...Thanks!

## Update :

Some related subject :

• What is ${\rm Li}(x)$? Commented Apr 14, 2020 at 10:15
• Commented Apr 14, 2020 at 10:31
• I am skeptical that this would have a nice closed form. Commented Apr 14, 2020 at 13:52
• @mrtaurho: Li(x) is the Offset Logarithmic Integral Function so corrected to li(x) the Logarithmic Integral Function. Result (3) applies to li(x) as referenced above. Commented Apr 14, 2020 at 16:39
• You could do the partial integration differently: $$\int_{0}^{1}\frac{x\operatorname{li}(x)}{x^2+1}dx=\left[\operatorname{li}(x)\cdot\left(\frac{1}{2}\ln(x ^2+1)-\frac{1}{2}\ln(2)\right)\right]_0^1-\int_{0}^{1}\frac{\ln(x^2+1)-\ln(2)}{2\ln(x)}dx$$ Then the first part is zero and it remains "only" the new integral. Commented Apr 17, 2020 at 5:07

Long Comment:

@L. Milla's integral can be written:

$$I=-\frac{1}{2}\int_0^1 \frac{\log \left(\frac{1}{2} \left((1-x)^2+1\right)\right)}{ \log (1-x)} \, dx\tag{1}$$

The first few terms series expansion approximation of $$\frac{1}{\log (1-x)}$$ are

$$\frac{1}{\log (1-x)}\approx-\frac{1}{x}+\frac{1}{2}+\frac{x}{12}+\frac{x^2}{24}+\frac{19 x^3}{720}+\frac{3 x^4}{160}+\frac{863 x^5}{60480}+...$$

https://oeis.org/A002206 and https://oeis.org/A002207 can be used to find the full series expansion, modifying the signs to fit the present purpose:

$$\frac{1}{\log (1-x)}=-\frac{1}{x}+\sum _{n=0}^{\infty } \frac{(-1)^{n-1} x^n }{n!}\left(\sum _{j=1}^{n+1} \frac{B_j S_n^{(j-1)}}{j}\right)$$

with $$B_j$$ being Bernoulli Numbers and $$S_n^{(j-1)}$$ being Stirling number of the first kind.

Quickly using Mathematica to integrate the first few terms of this expansion, perhaps results in a "generalised algebraic form" after the first two terms, rather than a closed form as such. i.e.

$$I\approx\left(\frac{\log ^2(2)}{8}-\frac{5 \pi ^2}{96}\right)+\left(\frac{4-\pi }{8}\right)$$

for the first two terms, with terms after that being of the general algebraic form

$$+\left(\frac{a-b \,\pi +c\, \log (2)}{d}\right)$$

with $$a$$,$$b$$,$$c$$ and $$d$$ being integers.

Not a complete answer, but too long for a comment.

First, let’s use the definition of the Logarithmic Integral Function. Then, let’s switch the order of integration.

$$I=\int_0^1 \frac{xli(x)}{1+x^2}dx=\int_0^1\int_0^x\frac{x}{(1+x^2)log(y)}dydx=\int_0^1\int_y^1\frac{x}{(1+x^2)log(y)}dxdy$$

$$I=\frac{1}{2} \int_0^1 \underbrace{\frac{log(2)-log(1+y^2)}{log(y)}}_{y=e^{-t}}dy=\frac{1}{2} \int_0^\infty \frac{log(1+e^{-2t})-log(2)}{t}e^{-t}dt$$

Now, let’s make use of Laplace Transform.

$$I=\frac{1}{2}\mathscr{L}\left[\frac{log(1+e^{-2t})-log(2)}{t}\right]_{s=1}=\frac{1}{4}\int_1^\infty \frac{\psi^{(0)}(\frac{s}{4}+\frac{1}{2})-\psi^{(0)}(\frac{s}{4}+1)}{s}ds$$

Rewritting the expression using some basic properties of Digamma Function. $$I=\frac{1}{4} \lim_{R\rightarrow \infty}\int_1^R \frac{2\psi^{(0)}(\frac{s}{2})-2log(2)-\psi^{(0)}(\frac{s}{4})-\psi^{(0)}(\frac{s}{4})-\frac{4}{s}}{s}ds$$

Now, let's integrate the functions that doesn't involve Digamma and let's apply some substitutions using the limit of each integral. $$I=-1-\frac{log(2)}{2}\lim_{R\rightarrow \infty}log(R)+\frac{1}{2}\lim_{R\rightarrow \infty}\int_1^R \underbrace{\frac{\psi^{(0)}(\frac{s}{2})}{s}}_{s\rightarrow2z}ds-\frac{1}{2}\lim_{R\rightarrow \infty}\int_1^R \underbrace{\frac{\psi^{(0)}(\frac{s}{4})}{s}}_{s\rightarrow4z}ds$$

$$I=-1-\frac{log(2)}{2}\lim_{R\rightarrow \infty}log(R)+\frac{1}{2}\lim_{R\rightarrow \infty}\int_{1/2}^{R/2} \frac{\psi^{(0)}(z)}{z}dz-\frac{1}{2}\lim_{R\rightarrow \infty}\int_{1/4}^{R/4} \frac{\psi^{(0)}(z)}{z}dz$$

$$I=-1-\frac{log(2)}{2}\lim_{R\rightarrow \infty}log(R)+\frac{1}{2}\lim_{R\rightarrow \infty}\int_{R/4}^{R/2} \frac{\psi^{(0)}(z)}{z}dz-\frac{1}{2}\int_{1/4}^{1/2} \frac{\psi^{(0)}(z)}{z}dz$$

Let's use the asymptotic expansion of $$\psi^{(0)}(z)$$ to evaluate the integral with limits at infinity and let's use Taylor Series to evaluate the integral with finite limits.

$$I=-1-\frac{log(2)}{2}\lim_{R\rightarrow \infty}log(R)+\frac{1}{2}\lim_{R\rightarrow \infty}\left[\frac{log^2(z)}{2}\right]^{R/2}_{R/4}+ \frac{1}{2}\left[-\frac{1}{s}+\gamma\log(s)+\sum_{k=1}^{\infty}\frac{\left(-1\right)^k\zeta\left(k+1\right)}{k}s^k\ \right]_{1/4}^{1/2}$$

$$I=-1-\frac{log(2)}{2}\lim_{R\rightarrow \infty}log(R)+\lim_{R\rightarrow\infty}\left[\frac{\log{\left(2\right)}}{2}\log{\left(R\right)}-\frac{3}{4}\log^2{\left(2\right)}\right]+\frac{1}{2}\left[-\left(2-4\right)-\left(\gamma\log{\left(2\right)}-2\gamma\log{\left(2\right)}\right)+\sum_{k=1}^{\infty}\frac{\left(-1\right)^k\zeta\left(k+1\right)}{k}\left(\frac{1}{2^k}-\frac{1}{2^{2k}}\right)\right]$$

$$I=-\frac{3}{4}\log^2{\left(2\right)}+\frac{\gamma\log{\left(2\right)}}{2}+\frac{1}{2}\sum_{k=1}^{\infty}{\frac{\left(-1\right)^k}{2^kk}\eta\left(k+1\right)}$$

$$\eta(z)$$ is the Dirichlet Eta Function.