These integrals came up in the process of finding solution to Vladimir Reshetnikov's problem. I wonder if there are closed-forms for the following integrals:
\begin{array}{1,1}
&[\text{1}] &\quad\int_0^1\frac{\operatorname{Li}_3(ax)}{1+2x}\ dx\\[12pt]
&[\text{2}] &\quad\int_0^1\frac{\operatorname{Li}_2(ax)\ln x}{1+2x}\ dx\\[12pt]
&[\text{3}] &\quad\int_0^1\frac{\ln(1-ax)\ln^2 x}{1+2x}\ dx
\end{array}
I have tried many substitutions, integration by parts, or differentiation under integral sign method, but without success so far. I do not need a complete or rigorous answer and your answer can be only Mathematica's or Maple's output since I don't have those software packages in my computer or links of related papers. I'd be grateful for any help you are able to provide.
[Math] Closed-forms for several tough integrals
calculusclosed-formdefinite integralsimproper-integralsintegration
Related Solutions
It is usually good to break up an integral into parts, so that each part has at most one "bad" feature.
For your third integral, break up into the integrals from $-\infty$ to $0$, and from $0$ to $\infty$.
Let's look at $\displaystyle \int_0^\infty \frac{e^x}{e^x+x^2}\,dx$. Informally, for large $x$, the $x^2$ term is utterly negligible in comparison with $e^x$. So for big $x$ our function is nearly $1$, and of course the integral will blow up.
To be more formal, divide top and bottom of our integrand by $e^x$. We get $$\dfrac{1}{1+\frac{x^2}{e^x}}.$$ Since $\lim_{x\to\infty} \frac{x^2}{e^x}=0$, for large enough $x$ we have $\frac{x^2}{e^x}\lt \frac{1}{2}$. So for large enough $x$, the integrand is $\gt \frac{2}{3}$.
The integral $\int_0^\infty \frac{2}{3}\,dx$ clearly diverges, so by Comparison so does $\displaystyle\int_0^\infty \frac{e^x}{e^x+x^2}\,dx$.
For your second integral, break up as you did. The ideas ou used are fine. We will prove that both integrals converge absolutely, meaning that if we take absolute values we still have convergence. By a standard theorem, absolute convergence implies convergence.
For the integral from $0$ to $1$, note that $|\sin(1/x)|\le 1$. Thus $$\left|\frac{\sin(1/x)}{x^{1/5}}\right|\le \frac{1}{x^{1/5}}.$$ By a standard theorem, $\int_0^1\frac{dx}{x^{1/5}}$ converges, so by Comparison our integral from $0$ to $1$ convrges absolutely, and hence converges.
For the integral from $1$ to $\infty$, I would suggest using the inequality $|\sin t|\le |t|$. For $x\ge 1$, this implies that $0\lt \sin(\pi/x)\lt \frac{\pi}{x}$.
Thus in our interval, our function is between $0$ and $\frac{\pi}{x^{6/5}}$. The rest is done by Comparison.
For the first integral, again the intuition is good. There could be improvement in the detail.
Break up as the integral from $0$ to $1$, plus the integral from $1$ to\infty$.
For the integral from $1$ to $\infty$, true, the $e^x$ term at the bottom will cush the top. Maybe you could use the Limit Comparison Test, comparing with $\frac{1}{e^{x/2}}$.
Or else observe that for large enough $x$ (and it happens pretty early) the bottom is bigger than $(1/2)e^{x}$. for large $x$, the top is $\lt e^{x/2}$. So the ratio, for large $x$, is $\lt \frac{2}{e^{x/2}}$.
For the part from $0$ to $1$, use the fact that $\lim_{x\to 0}\frac{e^x-1}{x}=1$, or more or less equivalently use the MacLaurin series for $e^x$ to get an idea of the size of $e^x-1$.
For $x$ close enough to $0$, $x/2\le e^x-1$. so our integrand is $\le \frac{2}{x^{1/2}}$, and by Comparison with a standard integral we have convergence.
This particular integral definitely looks simple. What changes when we try to evaluate it?
To start with, a simple substitution like $u=e^x, du=e^xdx$ goes like this:
$$\int\frac{dx}{1+2^x+3^x}=\int\frac {e^xdx}{e^x+e^{x(\ln 2+1)}+e^{x(\ln 3+1)}}=\int\frac{du}{u+u^{\ln 2+1}+u^{\ln 3+1}}$$
Already we have left the carefree world of simple functions behind, and this new function does not look easy... Perhaps a different approach would help:
$$3^x+2^x=(3^{\frac x2}+2^{\frac x2})^2-\sqrt{6^x}$$
No, that looks even worse.
This is part of the difficulty with integrals of this type; the common transformations that we are familiar with often fail to line up with certain problems that we would like to solve.
Since $1+2^x+3^x$ is strictly positive for $x\in\Bbb R$, we could try a substitution like $\cosh u=1+2^x+3^x,\sinh udu=(2^x\ln 2+3^x\ln 3)dx,$ but this fails to produce a viable substitution as well since we would be left with
$$\int\frac {\sinh u}{(2^x\ln 2+3^x\ln 3)\cosh u}du$$
as $\cosh u=1+2^x+3^x$ is not nicely soluble for $x,$ nor does any other obvious transformation present itself for placement as $f(u)=2^x\ln 2+3^x\ln 3$.
So indefinite integration by substitution is out (at least as far as the possibilities I am aware of), and integration by parts does not appear to yield any useful results. What about comparisons with definite integrals? What options are available here?
If the value of the integrals $\int\frac 1{1+3^x+3^x}dx=\int\frac 1{1+2\cdot 3^x}dx$ and $\int\frac 1{1+3^{\frac x2}+3^x}dx$ were known, this might be easier as we might be able to limit the possibilities...
In order to have a solvable integral, it is imperative that the denominator be factored fully, or at least to the point where partial fractions can take over and each part can be solved individually. The choice of $\int\frac 1{1+3^{\frac x2}+3^x}dx$ is one that is "near" to the original, and also happens to be cyclotomic, so that all the linear (complex) factors are well-known, and thus partial fractions is "easily" applicable.
In general, let $p_n(x)$ be the $n$th cyclotomic polynomial, and let $q_n(x)$ be the polynomial such that $p_n(x)\cdot q_n(x)=x^n-1.$ (For example, $p_3(2^x)=1+2^x+2^{2x}=1+2^x+4^x$, and $q_3(2^x)=2^x-1$.) Then the integral $\int\frac 1{p_n(\alpha^x)}dx$ can be expressed as
$$\int\frac{q_n(\alpha^x)}{\alpha^{nx}-1}dx=\int\frac{\sum\limits_{i=0}^{\deg(q)+1}a_i\alpha^{ix}}{\alpha^{nx}-1}dx=\sum_{i=0}^{\deg(q)+1}a_i\int\frac{\alpha^{ix}}{\alpha^{nx}-1}dx$$
With a substitution like $u=\alpha^x, du=(\ln\alpha)\alpha^xdx$ this becomes
$$\frac 1{\ln\alpha}\sum_{i=0}^{\deg(q)+1}a_i\int\frac{u^{i-1}}{u^n-1}du\tag 1$$
which is almost a direct translation to an integral with a cyclotomic polynomial; in fact, $(1)$ becomes
$$\frac 1{\ln\alpha}\int\frac{q_n(u)}{u(u^n-1)}du=\frac 1{\ln\alpha}\int\frac{q_n(u)}{u(u-1)\sum_{i=0}^{n-1}u^i}du=\frac1{\ln\alpha}\int\frac 1{up_n(u)}du\tag 2$$
After all that, we know that one of the integrals of interest above, namely, $\int\frac 1{1+3^{\frac x2}+3^x}dx$ can be written with transformation $u=(\sqrt 3)^x$ as
$$\frac2{\ln 3}\int\frac 1{u(u^2+u+1)}du\to\\x-\frac 1{\ln 3}\ln(3^x+3^{\frac x2}+1)-\frac 2{\sqrt 3\ln 3}\arctan\left(\frac 1{\sqrt 3}(2\cdot 3^{\frac x2}+1)\right)+c\tag 3$$
$$\int\frac 1{1+2\cdot 3^x}dx=x-\ln(2\cdot 3^x+1))+c\tag 4$$
Taking integration limits as $x\in[1,+\inf)$ from the question, we get the following values:
$$(3)|_1^\infty=-1+\frac{-\sqrt 3 \pi+2 \sqrt 3 \arctan(2+\frac 1{\sqrt 3})+3 \ln(4+\sqrt 3)}{\ln 27}\approx 0.2003338$$ $$(4)|_1^\infty=\frac{\ln\frac 76}{\ln 3}\approx 0.140314$$
This is only mildly useful as an upper and lower bound. While the value of the posted integral is definitely between these bounds, this is not a very accurate result and could certainly do with some improvement. Note that other integrals that fall outside these bounds (or do not supply sufficient information to be specified as bounds) include $\int\frac 1{1+2^{x+1}}dx,\int\frac 1{4^x+2^x+1}dx,\int\frac 1{4^x+1}dx,\int\frac 1{3^x+1}dx$.
Fortunately, the choice to integrate $\int\frac 1{3^x+3^{\frac x2}+1}dx$ suggests a route to solve the integral completely:
- Find a monotonic sequence of rationals $x_n=\frac ab$ whose limit is $\log_3 2$
- Find the value of $$I_n=\int\frac1{3^x+3^{x_n\cdot x}+1}dx$$ (this will be a three-term polynomial of some degree that will be factorable, possibly nicely depending on the sequence $x_n$ used)
- Find the limit $\lim_{n\to\infty}I_n$
This is not to say that these steps will be easy or computable in reasonable time frames...
Best Answer
Here are values of the integral $[3]$ for some specific values of the parameter $a$: $$\int_0^1\frac{\ln(1-x)\ln^2x}{1+2x}dx=2\operatorname{Li}_4\left(\frac12\right)-\operatorname{Li}_4\left(\frac13\right)-\operatorname{Li}_4\left(\frac23\right)-\frac14\operatorname{Li}_4\left(\frac14\right)\\-\frac{\ln^42}{12}-\frac{\ln^43}{12}+\frac16\ln2\cdot\ln^33+\frac{\pi^2}6\left(\ln2\cdot\ln3-\ln^22-\operatorname{Li}_2\left(\frac13\right)\right).\tag1$$
$$\int_0^1\frac{\ln(1+2x)\ln^2x}{1+2x}dx=\operatorname{Li}_4\left(\frac12\right)+\operatorname{Li}_4\left(\frac13\right)+\operatorname{Li}_4\left(\frac23\right)-\frac18\operatorname{Li}_4\left(\frac14\right)\\+\left(\operatorname{Li}_3\left(\frac13\right)+\operatorname{Li}_3\left(\frac23\right)\right)\ln3-\frac{11\pi^4}{360}-\frac{\ln^42}{24}-\frac{\ln^43}4\\+\frac{\pi^2}{12}\left(\ln^23-\ln^22\right)+\frac13\ln2\cdot\ln^33.\tag3$$