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5 votes
1 answer
193 views

Evaluating $\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{arsinh}{(ax)}\operatorname{arsinh}{(bx)}}{x}$ in terms of polylogarithms

Define the function $\mathcal{I}:\mathbb{R}^{2}\rightarrow\mathbb{R}$ by the definite integral $$\mathcal{I}{\left(a,b\right)}:=\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{arsinh}{\left(ax\right)}\...
David H's user avatar
  • 30.7k
2 votes
0 answers
68 views

Evaluating $\int\frac{\log(x+a)}{x}\,dx$ in terms of dilogarithms

As per the title, I evaluated $$\int\frac{\log(x+a)}{x}\,dx$$ And wanted to make sure my solution is correct, and if not, where I went wrong in my process. Here is my work. $$\int\frac{\log(x+a)}{x}\,...
Person's user avatar
  • 1,123
5 votes
1 answer
288 views

Closed forms of the integral $ \int_0^1 \frac{\mathrm{Li}_n(x)}{(1+x)^n} d x $

(This is related to this question). How would one find the closed forms the integral $$ \int_0^1 \frac{\mathrm{Li}_n(x)}{(1+x)^n} d x? $$ I tried using Nielsen Generalized Polylogarithm as mentioned ...
Anomaly's user avatar
  • 107
2 votes
1 answer
174 views

Calculate the integral of the given polylogarithm function? $\int_0^1\frac{\operatorname{Li}_ 4(x)}{1+x}dx=?$ [closed]

$$\int_0^1 \frac{\operatorname{Li}_2(-x)\operatorname{Li}_2(x)}{x}\,\mathrm dx=?$$ where $$\operatorname{Li}_2(-x)=\sum_{k=1}^{\infty}\frac{(-x)^k}{k^2}$$ for $$|x|>1$$ actually my goal is to edit ...
merve kaya's user avatar
3 votes
1 answer
244 views

How to solve $\int\frac{x\arctan x}{x^4+1}dx$ in a practical way

I need to evaluate the following indefinite integral for some other definite integral $$\int\frac{x\arctan x}{x^4+1}dx$$ I found that $$\int_o^\infty\arctan{(e^{-x})}\arctan{(e^{-2x})}dx=\frac{\pi G}{...
phi-rate's user avatar
  • 2,370
1 vote
0 answers
117 views

Closed-form for $\int_0^{a^2} \mathrm{Ei} (-s) \frac{1 - e^s}{s} ds$

In my partial answer to this question: Integral involving polylogarithm and an exponential, I arrive at the integral $$ \int_0^{a^2} \mathrm{Ei} (-s) \frac{1 - e^s}{s} ds , ~~~~ (\ast) $$ where $a \in ...
o0BlueBeast0o's user avatar
1 vote
1 answer
164 views

Integral involving product of dilogarithm and an exponential

I am interested in the integral \begin{equation} \int_0^1 \mathrm{Li}_2 (u) e^{-a^2 u} d u , ~~~~ (\ast) \end{equation} where $\mathrm{Li}_2$ is the dilogarithm. This integral arose in my attempt to ...
o0BlueBeast0o's user avatar
4 votes
2 answers
180 views

Prove $\int_0^1\frac{\text{Li}_2(-x^2)}{\sqrt{1-x^2}}\,dx=\pi\int_0^1\frac{\ln\left(\frac{2}{1+\sqrt{1+x}}\right)}{x}\,dx$

I managed here to prove $$\int_0^1\frac{\text{Li}_2(-x^2)}{\sqrt{1-x^2}}\,dx=\pi\int_0^1\frac{\ln\left(\frac{2}{1+\sqrt{1+x}}\right)}{x}\,dx$$ but what I did was converting the LHS integral to a ...
Ali Shadhar's user avatar
  • 25.8k
3 votes
2 answers
311 views

How can I evaluate $\int _0^1\frac{\operatorname{Li}_2\left(-x^2\right)}{\sqrt{1-x^2}}\:\mathrm{d}x$

I've been trying to find and prove that: $$\int _0^1\frac{\operatorname{Li}_2\left(-x^2\right)}{\sqrt{1-x^2}}\:\mathrm{d}x=\pi \operatorname{Li}_2\left(\frac{1-\sqrt{2}}{2}\right)-\frac{\pi }{2}\left(\...
mattsteiner64's user avatar
2 votes
1 answer
126 views

Closed form evaluation of a class of inverse hyperbolic integrals

Define the function $\mathcal{I}:\mathbb{R}_{>0}^{2}\rightarrow\mathbb{R}$ via the definite integral $$\mathcal{I}{\left(a,b\right)}:=\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{arsinh}{\left(ax\...
David H's user avatar
  • 30.7k
4 votes
3 answers
257 views

Arctan integral $ \int_{0}^{\infty}\frac{\arctan(x)}{x^{2}+k^{2}}$

Is there a closed form for the integral $$ \int_{0}^{\infty}\frac{\arctan(x)}{x^{2}+k^{2}}$$ for $\forall k \ge 1 $? Well, I was able to get the closed form for the case where $|k|\le1$, and it is of ...
Senna S's user avatar
  • 247
3 votes
1 answer
368 views

Challenging integral $I=\int_0^{\pi/2}x^2\frac{\ln(\sin x)}{\cos x}dx$

My friend offered to solve this integral. $$I=\int_0^{\pi/2}x^2\frac{\ln(\sin x)}{\cos x}dx=\frac{\pi^4}{32}-{4G^2} $$ Where G is the Catalan's constant. $$I=\int _0^{\infty }\frac{\arctan ^2\left(u\...
user178256's user avatar
  • 5,507
4 votes
1 answer
286 views

Evaluate $\int^1_0 x^a (1-x)^b \operatorname{Li}_2 (x)\, \mathrm dx$

For what $a,b$ the integral $$\int^1_0 x^a(1-x)^b\operatorname{Li}_2 (x)\, \mathrm dx$$ has a closed form solution? I tried to solve it by expanding dilogarithm function, or by reducing it to linear ...
Machinato's user avatar
  • 2,903
3 votes
0 answers
184 views

How to simplify this polylog expression $\operatorname{Li}_4\left(\frac{z-1}{z}\right)$?

Evaluating this integral in Mathematica $$i_2(z)=\int_0^z \frac{\log ^2(x) \log (1-x)}{1-x} \, dx\tag{1}$$ returns a mixture of polylogs up to order 4 and several log-terms. In the region of ...
Dr. Wolfgang Hintze's user avatar
7 votes
3 answers
615 views

Closed-forms for the integral $\int_0^1\frac{\operatorname{Li}_n(x)}{1+x}dx$?

(This is related to this question.) Define the integral, $$I_n = \int_0^1\frac{\operatorname{Li}_n(x)}{1+x}dx$$ with polylogarithm $\operatorname{Li}_n(x)$. Given the Nielsen generalized polylogarithm ...
Tito Piezas III's user avatar

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