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How can I evaluate

$$ \int_{0}^{1} \frac{\ln(x+\sqrt{1-x^2})}{\sqrt{1+x^2}} \, \mathrm{d}x $$

U-substitution has not worked for me. Integration by parts, Differentiation under integral sign, Mathematica is not coming up with a solution either.

Is there a closed form for this integral?

Thank you kindly for your help and time.

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    $\begingroup$ This is an intriguing problem, but why do you need to do it? Where did it come up? $\endgroup$ – Ninad Munshi Aug 29 at 1:08
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    $\begingroup$ Most probabily he means $\displaystyle \int^{1}_{0}\frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}dx$ $\endgroup$ – jacky Aug 29 at 2:56
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    $\begingroup$ @jacky I don't want to speak for the OP, but I don't think so. Mathematica was able to easily solve that integral. $\endgroup$ – Varun Vejalla Aug 29 at 3:02
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    $\begingroup$ Mathrmatica solved this in a second: $\frac12 \text{arcsinh}^2(1)$ $\endgroup$ – Oldboy Aug 29 at 6:57
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    $\begingroup$ @Oldboy That is for jacky's integral, not the one posted in the question. $\endgroup$ – Varun Vejalla Aug 29 at 7:15
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$$\small \int_0^1 \frac{\ln(x+\sqrt{1-x^2})}{\sqrt{1+x^2}}dx=\frac{5}{4}\ln^2 2-\frac94\ln 2\ln(1+\sqrt 2)+\frac98\ln^2(1+\sqrt 2)-\frac{3\pi^2}{16}$$ $$\small +\ln(1+\sqrt 3)\left(\ln(1+\sqrt 2)-\frac12\ln 2\right)+\ln\left(\frac{1}{1-\sqrt 3}\right)\left(\frac12\ln 2 -\ln(1+\sqrt 2)\right)$$ $$\small +\frac12\operatorname{Li}_2\left(1-\frac{1}{\sqrt 2}\right)+\frac34\operatorname{Li}_2\left(\sqrt 2-1\right)+\frac14\operatorname{Li}_2\left(1-\sqrt 2\right)-\frac32\operatorname{Li}_2\left(-(1+\sqrt 2)\right)+\frac12\operatorname{Li}_2\left(2-\sqrt 2\right)$$ $$\small+\frac12\operatorname{Li}_2\left(\frac34\right)-\frac12\operatorname{Li}_2\left(3\left(1-\frac{1}{\sqrt 2}\right)\right)+\frac38\operatorname{Li}_2\left(-(1+\sqrt 2)^2\right)-\frac18\operatorname{Li}_2\left(-(\sqrt 2-1)^2\right)+\frac12\operatorname{Li}_2\left((\sqrt 2-1)^2\right)$$ $$\small-\operatorname{Li}_2\left(\frac12(3-\sqrt 3)\right)+\operatorname{Li}_2\left(\left(1-\frac{1}{\sqrt 2}\right)(3-\sqrt 3)\right)-\operatorname{Li}_2\left(\frac12(3+\sqrt 3)\right)+\operatorname{Li}_2\left(\left(1-\frac{1}{\sqrt 2}\right)(3+\sqrt 3)\right)$$


To show the result from above we'll start by splitting into two integrals. $$\int_0^1 \frac{\ln(x+\sqrt{1-x^2})}{\sqrt{1+x^2}}dx=\int_0^1\frac{\ln\left(1+\frac{\sqrt{1-x^2}}{x}\right)}{\sqrt{1+x^2}}dx+\int_0^1 \frac{\ln x}{\sqrt{1+x^2}}dx=I+J$$ Starting with the first integral we'll first make some substitutions, namely: $x=\cos t$; $\tan t = x$ and in the end to get rid of the square roots we'll use an Euler substitution: $\sqrt{2+x^2}-x=t\Leftrightarrow x=\frac{2-t^2}{2t}$. $$\small I=\int_0^\frac{\pi}{2}\frac{\sin x\ln(1+\tan x)}{\sqrt{1+\cos^2 x}}dx\overset{\tan x\to x}=\int_0^\infty\frac{x\ln(1+x)}{(1+x^2)\sqrt{2+x^2}}dx\overset{x\to \frac{2-x^2}{2x}}=\int_0^\sqrt 2\ln\left(\frac{2+2x-x^2}{2x}\right)\frac{4-2x^2}{4+x^4}dx$$ $$=-\ln 2\ln(1+\sqrt 2)+\int_0^\sqrt 2\left(\ln(2+2x-x^2)-\ln x\right)\left(\frac{1-x}{1+(1-x)^2}+\frac{1+x}{1+(1+x)^2}\right)dx$$ Now we just need to split everything into four integrals and evaluate them in order to find $I$. Mostly we will just use directly the following result:

$$\int \frac{\ln(a+bx)}{1+x}dx\overset{1+x=t}=\int \frac{\ln(a-b+bt)}{t}dt=\int\frac{\ln(a-b)+\ln\left(1-\frac{b}{b-a}t\right)}{t}dt$$ $$\overset{\frac{b}{b-a}t=y}=\ln(a-b)\ln t+\int \frac{\ln(1-y)}{y}dy=\ln(a-b)\ln(1+x)-\operatorname{Li}_2\left(\frac{b(1+x)}{b-a}\right)+C\tag 1$$


Let's start with the easiest one.

$$K=\int_0^\sqrt 2 \frac{(1-x)\ln(2+2x-x^2)}{1+(1-x)^2}dx\overset{1-x\to x}=\int_{1-\sqrt 2}^1\frac{x\ln(3-x^2)}{1+x^2}dx$$ $$\overset{x^2\to x}=\frac12\int_{(1-\sqrt 2)^2}^1\frac{\ln(3-x)}{1+x}dx=\ln 2\ln(1+\sqrt 2)-\frac14\ln^2 2-\frac{\pi^2}{24}+\frac12\operatorname{Li}_2\left(1-\frac{1}{\sqrt 2}\right)$$ Where the result mentioned in $(1)$ was used alongside $\operatorname{Li}_2\left(\frac12\right)=\frac{\pi^2}{12}-\frac12\ln^2 2$.


For the next integral we will consider it's "sister" integral and evaluate them combined, e.g. $A+B$ and $A-B$ then extract it as $A=\frac12((A+B)+(A-B))$.

$$A=\int_0^\sqrt 2\frac{(1-x)\ln x}{1+(1-x)^2}dx=\int_{1-\sqrt 2}^1\frac{x\ln(1-x)}{1+x^2}dx;\quad B=\int_{1-\sqrt 2}^1\frac{x\ln(1+x)}{1+x^2}dx$$ $$A+B=\int_{1-\sqrt 2}^1\frac{x\ln(1-x^2)}{1+x^2}dx=\frac12\int_{(1-\sqrt 2)^2}^1\frac{\ln(1-x)}{1+x}dx=$$ $$=\frac34\ln 2\ln(1+\sqrt 2)-\frac14\ln^2 2-\frac12\ln^2(1+\sqrt 2)-\frac12\operatorname{Li}_2(\sqrt 2-1)$$ $$A-B=\int_{1-\sqrt 2}^1\frac{x\ln\left(\frac{1-x}{1+x}\right)}{1+x^2}dx=\int_0^{1+\sqrt 2}\ln x\left(\frac{1}{1+x}-\frac{x}{1+x^2}\right)dx$$ $$=\frac12\ln^2(1+\sqrt 2)-\frac14\ln 2\ln(1+\sqrt 2)+\operatorname{Li}_2\left(-(1+\sqrt 2)\right)-\frac14\operatorname{Li}_2\left(-(1+\sqrt 2)^2\right)$$ Above, the second integral reduces to the first one integral after the substitution $x^2\to x$ and $(1)$ is applicable with $a=0,b=1$. In the end we obtain: $$A=\frac14\ln 2\ln(1+\sqrt 2)-\frac18\ln^2 2-\frac14\operatorname{Li}_2\left(\sqrt 2-1\right)+\frac12\operatorname{Li}_2\left(-(1+\sqrt 2)\right)-\frac18\operatorname{Li}_2\left(-(1+\sqrt 2)^2\right)$$


This one follows exactly the same approach as above. $$C=\int_0^\sqrt 2\frac{(1+x)\ln x}{1+(1+x)^2}dx=\int_1^{1+\sqrt 2}\frac{x\ln(x-1)}{1+x^2}dx;\quad D=\int_1^{1+\sqrt 2}\frac{x\ln(x+1)}{1+x^2}dx$$ $$C+D=\int_1^{1+\sqrt 2}\frac{x\ln(x^2-1)}{1+x^2}dx=\frac12\int_1^{(1+\sqrt 2)^2}\frac{\ln(x-1)}{1+x}dx=$$ $$=\frac34\ln 2\ln(1+\sqrt 2)+\frac14\ln^2 2+\frac14\ln^2(1+\sqrt 2)-\frac{\pi^2}{12}-\frac12\operatorname{Li}_2(1-\sqrt 2)$$ $$C-D=\int_1^{1+\sqrt 2}\frac{x\ln\left(\frac{x-1}{x+1}\right)}{x^2+1}dx=\int_0^{\sqrt 2-1}\ln x\left(\frac{1}{1-x}+\frac{x}{1+x^2}\right)dx$$ $$=-\frac14\ln 2\ln(1+\sqrt 2)-\frac12\ln^2(1+\sqrt 2)+\frac14\operatorname{Li}_2\left(-(\sqrt 2-1)^2\right)-\operatorname{Li}_2\left(\sqrt 2-1\right)$$ $$\Rightarrow C=\frac14\ln 2\ln(1+\sqrt 2)+\frac18\ln^2 2-\frac18\ln^2(1+\sqrt 2)-\frac{\pi^2}{24} $$ $$+\frac18\operatorname{Li}_2\left(-(\sqrt 2-1)^2\right)-\frac12\operatorname{Li}_2\left(\sqrt 2-1\right)-\frac14\operatorname{Li}_2\left(1-\sqrt 2\right)$$


Finally we have just one integral left in order to finish with $I$, for this one we'll split again into four integrals after a few substitutions.

$$Q=\int_0^\sqrt 2 \frac{(1+x)\ln(2+2x-x^2)}{1+(1+x)^2}dx\overset{1+x\to x}=\int_{1}^{1+\sqrt 2}\frac{x\ln(4x-1-x^2)}{1+x^2}dx$$ $$\overset{x\to \frac{1-x}{1+x}}=\int_{1-\sqrt 2}^0\ln\left(\frac{2(1-3x^2)}{(1+x)^2}\right)\left(\frac{1}{1+x}-\frac{x}{1+x^2}\right)dx$$ $$=\frac14\ln^2 2+\frac12\ln 2\ln(1+\sqrt 2)-Q_1-2Q_2+Z+2X$$


$$Q_1=\int_{1-\sqrt 2}^0\frac{x\ln(1-3x^2)}{1+x^2}dx=-\int_0^{(1-\sqrt 2)^2}\frac{\ln(1-3x)}{1+x}dx$$ $$=\ln 2\ln(1+\sqrt 2)-\frac32\ln^2 2-\frac12\operatorname{Li}_2\left(\frac34\right)+\frac12\operatorname{Li}_2\left(3\left(1-\frac{1}{\sqrt 2}\right)\right)$$

$$Q_2=\int_{1-\sqrt 2}^0 \frac{\ln(1+x)}{1+x}dx=\frac12\ln 2\ln(1+\sqrt 2)-\frac12\ln^2(1+\sqrt 2)-\frac18\ln^2 2$$


$$X=\int_{1-\sqrt 2}^0\frac{x\ln(1+x)}{1+x^2}dx;\quad Y=\int_{1-\sqrt 2}^0\frac{x\ln(1-x)}{1+x^2}dx$$ $$X+Y=\int_{1-\sqrt 2}^0\frac{x\ln(1-x^2)}{1+x^2}dx=-\frac12\int_0^{(1-\sqrt 2)^2}\frac{\ln(1-x)}{1+x}dx$$ $$=\frac12\ln 2\ln(1+\sqrt 2)-\frac12\ln^2 2-\frac{\pi^2}{24}+\frac12 \operatorname{Li}_2(2-\sqrt 2)$$ $$X-Y=\int_{1-\sqrt 2}^0\frac{x\ln\left(\frac{1+x}{1-x}\right)}{1+x^2}dx\overset{x\to\frac{1-x}{1+x}}=\int_1^{1+\sqrt 2}\ln x\left(\frac{x}{1+x^2}-\frac{1}{1+x}\right)dx$$ $$=\frac14\ln 2\ln(1+\sqrt 2)-\frac12\ln^2(1+\sqrt 2)-\frac{\pi^2}{16}-\operatorname{Li}_2(-(1+\sqrt 2))+\frac14\operatorname{Li}_2(-(1+\sqrt 2)^2)$$ And similarly to how we found $A$ and $C$ we will extract $X$ from $X=\frac12\left((X+Y)+(X-Y)\right)$. $$\Rightarrow X=\frac38\ln 2\ln(1+\sqrt 2)-\frac14\ln^2(1+\sqrt 2)-\frac14\ln^2 2-\frac{5\pi^2}{96}$$ $$+\frac14 \operatorname{Li}_2(2-\sqrt 2)-\frac12\operatorname{Li}_2(-(1+\sqrt 2))+\frac18\operatorname{Li}_2(-(1+\sqrt 2)^2)$$


$$Z=\int_{1-\sqrt 2}^0\frac{\ln(1-3x^2)}{1+x}dx=\int_{1-\sqrt 2}^0\frac{\ln(1-\sqrt 3x)}{1+x}dx+\int_{1-\sqrt 2}^0\frac{\ln(1+\sqrt 3x)}{1+x}dx$$ $$=\ln(1+\sqrt 3)\left(\ln(1+\sqrt 2)-\frac12\ln 2\right)-\operatorname{Li}_2\left(\frac12(3-\sqrt 3)\right)+\operatorname{Li}_2\left(\left(1-\frac{1}{\sqrt 2}\right)(3-\sqrt 3)\right)$$ $$+\ln\left(\frac{1}{1-\sqrt 3}\right)\left(\frac12\ln 2 -\ln(1+\sqrt 2)\right)-\operatorname{Li}_2\left(\frac12(3+\sqrt 3)\right)+\operatorname{Li}_2\left(\left(1-\frac{1}{\sqrt 2}\right)(3+\sqrt 3)\right)$$


And there's just one integral left, the second one from the very beginning. $$J=\int_0^1 \frac{\ln x}{\sqrt{1+x^2}}dx\overset{x\to \frac{1-x^2}{2x}}=\int_{\sqrt 2-1}^1 \frac{\ln(1-x^2)-\ln 2-\ln x}{x}dx$$ $$=\frac12\ln^2(1+\sqrt 2)-\ln 2\ln(1+\sqrt 2)-\frac{\pi^2}{12}+\frac12\operatorname{Li}_2((\sqrt 2-1)^2)$$ Above follows as after the substitution $x^2= t$ we get $\int \frac{\ln(1-x^2)}{x}dx=-\frac12\operatorname{Li}_2(x^2)+C$.
Finally combining every result as $I+J=-\ln 2\ln(1+\sqrt 2)+K-A-C+Q+J$ gives the announced result.

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  • $\begingroup$ check by differentiation $\endgroup$ – C Squared Aug 30 at 23:39
  • $\begingroup$ You miscalculated $X$. I think this may have been partly because you defined $Y$ the same as $X$. The correct value (obtained with Mathematica) for $X$ is $$\frac{5\pi^2}{96} - \frac{\ln^2(2)}{8} - \frac{\ln^2(2-\sqrt{2})}{2} - \frac{1}{2}\left( \text{Li}_2\left((1+i)\left(1-\frac{1}{\sqrt{2}}\right)\right) + \text{Li}_2\left((1-i)\left(1-\frac{1}{\sqrt{2}}\right)\right) \right)$$ I don't know if this can be simplified further. $\endgroup$ – Varun Vejalla Aug 31 at 1:47
  • $\begingroup$ @VarunVejalla Thank you, for some reason a wild $\frac{\pi^2}{16}$ appeared there, even though there was another $\pi^2$ term. Numerically now it matches (wolfram link) and I've also corrected the $Y$ integral. If you can please let me know if anything else doesn't seem right. $\endgroup$ – Zacky Aug 31 at 6:32
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    $\begingroup$ I came up with same strategy but fail to solve the first integral. This is beautiful solution, (+1). $\endgroup$ – Naren Aug 31 at 6:59
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To calculate this integral I'll use series expansion $$\frac 1 {\sqrt{1+x^2}}=\sum_{n=0}^{\infty }\frac {(-1)^n} {2^{2n}}\binom{2n}{n}x^{2n}$$ for $|x|\le1$

$$I=\int_{0}^{1}\sum_{n=0}^{\infty }\frac {(-1)^n} {2^{2n}}\binom{2n}{n}x^{2n} \ln\left(x+\sqrt{1-x^2}\right)dx$$

By dominated convergence

$$I=\sum_{n=0}^{\infty }\frac {(-1)^n} {2^{2n}}\binom{2n}{n} \int_{0}^{1} \ln\left(x+\sqrt{1-x^2}\right)x^{2n} dx$$

Let $$ J=\int_{0}^{1} \ln\left(x+\sqrt{1-x^2}\right)x^{2n}$$

Now, Let $x=\cos\theta$

$$\implies J=\int_{0}^{\fracπ2}\ln\left(\cos\theta+\sin\theta\right)\left(\cos^{2n}\theta\right) (\sin\theta) d\theta$$

$$ \implies J=\frac12 \int_{0}^{\fracπ2}\ln\left(1+\sin2\theta\right)\left(\cos^{2n}\theta\right)\left(\sin\theta\right) d\theta$$

$$ \implies J=\frac12 \int_{0}^{\fracπ2}\left(\cos^{2n}\theta\right) \left(\sin\theta\right) \sum_{k=1}^{\infty }(-1)^{k-1}\frac {\sin^k 2\theta}{k} d\theta$$

$$ \implies J=\frac12 \int_{0}^{\fracπ2}\left(\cos^{2n}\theta\right) \left(\sin\theta\right) \sum_{k=1}^{\infty }(-1)^{k-1}\frac {2^k \left(\sin^k \theta \right)\left(\cos^k\theta\right)}{k} d\theta$$

By dominated convergence

$$J= \sum_{k=1}^{\infty }\frac {(-1)^{k-1} 2^{k-1}}{k}\int_{0}^{\fracπ2}\left(\cos^{2n+k}\theta\right) \left(\sin^{k+1}\theta\right) d\theta$$

Using $$\int_{0}^{\fracπ2}\left(\sin^m\theta\right) \left(\cos^n\theta\right)d\theta=\frac{\Gamma\left(\frac{n+1}2\right) \Gamma\left(\frac{m+1}2\right)}{2 \Gamma\left(\frac{m+n+2}2\right)}$$

$$J=\sum_{k=1}^{\infty }\frac{(-1)^{k-1} 2^{k-2}}{k}\frac{\Gamma\left(\frac{k+2}2\right) \Gamma\left(\frac{2n+k+1}2\right)}{ \Gamma\left(\frac{2n+2k+3}2\right)}$$

On substituting $J$ in orignal integral, we get

$$I=\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}\frac {(-1)^{(n+k-1)}}{2^{(2n-k+2)}k}\binom{2n}{n}\frac{\Gamma\left(\frac{k+2}2\right) \Gamma\left(\frac{2n+k+1}2\right)}{ \Gamma\left(\frac{2n+2k+3}2\right)}$$

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    $\begingroup$ Nice job on getting that double sum (+1)! By the way, you can put a \ before sin, cos, ln, etc to get it formatted nicely. E.g. using \cos instead of cos yields $\cos$ instead of $cos$. What I like to do to get nice formatting is type in Desmos, then copy that MathJax. $\endgroup$ – Varun Vejalla Aug 29 at 21:13
  • $\begingroup$ @Varun Vejalla Thanks $\endgroup$ – Paras Aug 29 at 21:29
  • $\begingroup$ Hostmath is also a quite nice tool to use for visual LaTeX editing. $\endgroup$ – Cade Reinberger Sep 1 at 19:41

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