Is there a nice/clever method to derive a general closed form for:

$$\displaystyle \int_0^1 \frac{\ln(1+x^a)}{1+x}dx, \;\ a>1\quad?$$

I thought maybe start with differentiating w.r.t. $a$.

This gives $\displaystyle \int_0^1 \frac{x^{a}\ln(x)}{(1+x^{a})(1+x)}dx$.

Maybe even use $\ln(1+x^{a})=\int_0^{x^{a}}\frac{1}{1+t}dt$ and/or series somehow.

But, now is there some way to link it to digamma, incomplete beta function, polylog, or some other advanced function?.

I just got to wondering about this one. If a general from can be derived, it would be

handy for many values of $a$. Thanks very much.

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Expanding $\log(1+x^a) = \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k} x^{a k}$ and integrating term-wise: $$\int_0^1 \frac{\log(1+x^a)}{1+x} \mathrm{d} x = \sum_{k=1}^\infty \frac{(-1)^{k-1}}{2 k} \left( \psi\left( \frac{a k}{2} + 1 \right) - \psi\left( \frac{a k}{2} + \frac{1}{2} \right) \right)$$

Integrate by parts: $$\log(1+x^a) \mathrm{d}\log(1+x) = \mathrm{d} \left( \log(1+x) \log(1+x^a) \right) - \log(1+x) \mathrm{d} \log(1+x^a)$$ Therefore $$\begin{eqnarray} \int_0^1 \frac{\log(1+x^a)}{1+x} \mathrm{d} x &=& \log^2(2) - a \int_0^1 \frac{\log(1+x) x^{a-1}}{1+x^a} \mathrm{d} x \\ &=& \log^2(2) - \sum_{k=1}^\infty \frac{(-1)^{k-1}}{2 k} \left( \psi\left( \frac{k}{2 a} + 1 \right) - \psi\left( \frac{k}{2a } + \frac{1}{2} \right) \right) \end{eqnarray}$$

Notice that this implies, that for $a=1$, the result is $\frac{1}{2} \log^2(2)$.

These sums can not be evaluated in closed forms, I am afraid, unless $a$ is a rational number.

Notice that $a$ need not be greater than $1$ in order to assure convergence of the integral. It can be any real number.

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Thanks, Sasha. Very nice. I thought it could probably be related to the digamma in some manner. The reason I asked was that I saw where someone had asked to prove that $\displaystyle \int_{0}^{1}\frac{ln(1+x^{2+\sqrt{3}})}{1+x}dx=\frac{{\pi}^{2}}{12}(1-\sqrt{3})+‌​ln(2)ln(1+\sqrt{3})$. I thought maybe a general form would be nice. But, I do not believe $2+\sqrt{3}$ is rational, so perhaps another method may be in order. – Cody Dec 14 '11 at 19:30
@Cody Thanks, this is a nice identity. I would like to see its proof as well. – Sasha Dec 14 '11 at 20:21

There's an interesting series expansion around $a=0$, which I think must be an asymptotic series:

$$\int_0^1 \frac{\ln(1+x^a)}{1+x}\ dx = \ln(2)^2 - \frac{\pi^2}{24} a + \sum_{k=1}^\infty \left(4^k + 4^{-k} - 2\right) \frac{B_{2k}\ \zeta(1+2k)}{2k} a^{2k}$$

where $B_{2k}$ are Bernoulli numbers.

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It would follow from the series in terms of $\psi(k/(2a)+ \cdot )$ in my post by using asymptotic series for digammas. – Sasha Dec 14 '11 at 20:55

I have noticed some other interesting relationships with log integrals and $2+\sqrt{3}$. $\tan(\frac{\pi}{12})=2-\sqrt{3}$. For instance, $\displaystyle \frac{3}{2}\int_{2+\sqrt{3}}^{\infty}\frac{\ln(x)}{1+x^{2}}\mathrm{d}x=\text{Catalan constant}$. There is an interesting identity that can be shown using the addition formula for tangent: $\displaystyle 2\int_{0}^{\frac{\pi}{12}}\ln(\tan(3x))dx=\int_{0}^{\frac{\pi}{12}}\ln(\tan(x))\mathrm{d}x$. Let $y=3x$ and we get $\displaystyle\frac{-3}{2}\int_{0}^{\frac{\pi}{12}}\ln(\tan(x))\mathrm{d}x=-\int_{0}^{\frac{\pi}{4}}\ln(\tan(y))\mathrm{d}y$. By letting $y=1/x$ we can also show $\displaystyle \frac{3}{2}\int_{2+\sqrt{3}}^{\infty}\frac{\ln(x)}{1+x^{2}}\mathrm{d}x$. There is that $2+\sqrt{3}$ again. I have a feeling something is in there somewhere.

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Sorry, Sasha, for posting my thread out of order. That was unintentional. – Cody Dec 15 '11 at 14:23