Evaluate $\int_0^1\frac{\ln{(x^2+1)}}{x+1}dx$

Question

Any idea on how to solve the following definite integral?

$$\int_0^1\frac{\ln{(x^2+1)}}{x+1}dx$$

I have tried to parameterize the integral like $\ln{(a^2x^2+1)}$ or $\ln{(x^2+a^2)}$, which don't seem to work.

Answer 1

\begin{equation} I = \int_0^1 \frac{\ln\left(x^2 + 1\right)}{x + 1}\:dx \end{equation}

Here let:

\begin{equation} I(t) = \int_0^1 \frac{\ln\left(tx^2 + 1\right)}{x + 1}\:dx \end{equation}

We observe that $I = I(1)$ and $I(0) = 0$. Thus,

\begin{align} I'(t) &= \int_0^1 \frac{x^2}{\left(tx^2 + 1\right)\left(x + 1\right)}\:dx = \frac{1}{t + 1}\int_0^1 \left[\frac{1}{x + 1} + \frac{x}{tx^2 + 1} - \frac{1}{tx^2 + 1} \right]\:dx \\ &= \frac{1}{t + 1}\bigg[\ln(x + 1) + \frac{1}{2t}\ln\left(tx^2 + 1 \right) + \frac{1}{\sqrt{t}}\arctan\left(\sqrt{t}x\right) \bigg]_0^1 \\ &=\frac{1}{t + 1} \bigg[\ln(2) + \frac{1}{2t}\ln\left(t + 1\right) + \frac{1}{\sqrt{t}}\arctan\left(\sqrt{t}\right)\bigg] \end{align}

Thus,

\begin{align} I(t) &= \int\frac{1}{t + 1} \bigg[\ln(2) + \frac{1}{2t}\ln\left(t + 1\right) + \frac{1}{\sqrt{t}}\arctan\left(\sqrt{t}\right)\bigg]\:dt \\ &= \int\frac{\ln(2)}{t + 1}\:dt + \int \frac{1}{2t\left(t + 1\right)}\ln\left(t + 1\right)\:dt + \int \frac{1}{\sqrt{t}\left(t + 1\right)}\arctan\left(\sqrt{t}\right)\:dt \\ &= I_1 + I_2 + I_3 \end{align}

Solve each individually:

For $I_1$:

\begin{equation} I_1 = \int\frac{\ln(2)}{t + 1}\:dt = \ln(2)\ln(t + 1) \end{equation}

For $I_2$:

\begin{align} I_2 &= \int\frac{1}{2t\left(t + 1\right)}\ln\left(t + 1\right)\:dt = \frac{1}{2}\left[ \int \left(\frac{1}{t} - \frac{1}{t + 1} \right)\ln(t + 1)\:dt\right]\\ & =\frac{1}{2} \left[\int \frac{\ln(t + 1)}{t}\:dt - \int \frac{\ln(t + 1)}{t + 1}\:dt \right] \\ &= \frac{1}{2} \left[-\operatorname{Li}_{2}(-t) - \frac{1}{2}\ln^2(t + 1) \right] \end{align}

Where $\operatorname{Li}_{2}(t)$ is Dilogarithm function.

For $I_3$:

\begin{equation} \int \frac{1}{\sqrt{t}\left(t + 1\right)}\arctan\left(\sqrt{t}\right)\:dt = \arctan^2\left(\sqrt{t}\right) \end{equation}

Combining we arrive at:

\begin{equation} I(t) = \ln(2)\ln(t + 1) + \frac{1}{2} \left[-\operatorname{Li}_{2}(-t) - \frac{1}{2}\ln^2(t + 1) \right] + \arctan^2\left(\sqrt{t}\right) + C \end{equation} Where $C$ is the constant of integration.

Using $I(0) = 0$ we see that $C = 0$. Thus

\begin{equation} I(t) = \ln(2)\ln(t + 1) + \frac{1}{2} \left[-\operatorname{Li}_{2}(-t) - \frac{1}{2}\ln^2(t + 1) \right] + \arctan^2\left(\sqrt{t}\right) \end{equation}

And finally:

\begin{align} I = I(1) &= \ln(2)\ln(1 + 1) + \frac{1}{2} \left[-\operatorname{Li}_{2}(-1) - \frac{1}{2}\ln^2(1 + 1) \right] + \arctan^2\left(\sqrt{1}\right) \\ &=\ln^2(2) + \frac{1}{2} \left[-\frac{\pi^2}{12} - \frac{1}{2}\ln^2(2)\right] + \frac{\pi^2}{16} \\ &= \frac{3}{4}\ln^2(2) - \frac{\pi^2}{48} \end{align}

Answer 2

The trick to this integral is a round of integration by parts followed by a direct application of Feynman's Trick. To wit, denote the integral$$\mathfrak{I}=\int\limits_0^1\mathrm dx\,\frac {\log(1+x^2)}{1+x}$$ Now integrate $\mathfrak I$ by parts letting $u=\log(1+x^2)$ and $v=\log(1+x)$. What follows is$$\begin{align*}\mathfrak{I} & =\log(1+x)\log(1+x^2)\,\Biggr\rvert_0^1-2\int\limits_0^1\mathrm dx\,\frac {x\log(1+x)}{1+x^2}\\ & =\log^22-2\int\limits_0^1\mathrm dx\,\frac {x\log(1+x)}{1+x^2}\end{align*}$$

The remaining integral can be evaluated using Feynman's Trick. Insert a parameter $a$ inside the natural logarithm and let $F(a)$ denote the integral. Note that $F(1)$ is what we seek.$$F(a)=\int\limits_0^1\mathrm dx\,\frac {x\log(1+ax)}{1+x^2}$$Proceeding normally, we differentiate with respect to $a$ and separate the remainder into partial fractions to get$$\begin{align*}F'(a) & =\int\limits_0^1\mathrm dx\,\frac {x^2}{(1+ax)(1+x^2)}\\ & =\int\limits_0^1\frac {\mathrm dx}{1+ax}-\int\limits_0^1\frac {\mathrm dx}{(1+ax)(1+x^2)}\end{align*}$$The first integral is trivial$$\int\limits_0^1\frac {\mathrm dx}{1+ax}=\frac {\log(1+a)}a$$The second integral can be evaluated again by separating the fraction and integrating from zero to one. In short, one arrives at the ugly result$$\int\limits_0^1\frac {\mathrm dx}{(1+ax)(1+x^2)}=\frac {a\log(1+a)}{1+a^2}+\frac {a\log 2}{2a^2+2}-\frac {\pi}{4a^2+4}$$Putting everything together, then$$F'(a)=\frac {\log(1+a)}a-\frac {a\log(1+a)}{1+a^2}+\frac {a\log 2}{2a^2+2}-\frac {\pi}{4a^2+4}$$

To arrive back at $F(1)$, we exploit a nifty little trick by instead of taking the indefinite integral of each term, integrate with respect to $a$ from zero to one. The middle term then simply becomes $F(1)$ and the others can be written in terms of the logarithm. Doing all the work, one gets that$$\begin{align*}F(1) & =\frac 12\int\limits_0^1\mathrm da\,\frac {\log(1+a)}a+\frac 14\log 2\int\limits_0^1\mathrm da\,\frac a{1+a^2}-\frac {\pi}8\int\limits_0^1\frac {\mathrm da}{1+a^2}\\ & =\frac {\pi^2}{24}+\frac 18\log^22-\frac {\pi^2}{32}\\ & \,\,\color{red}{=\frac 18\log^22+\frac {\pi^2}{96}}\end{align*}$$

Hence, multiplying $F(1)$ above by two and substituting everything back in, we arrive at the answer as the comments$$\int\limits_0^1\mathrm dx\,\frac {\log(1+x^2)}{1+x}\color{blue}{=\frac 34\log^22-\frac {\pi^2}{48}}$$

Answer 3

Not a full answer (yet)

I'll try to provide a solution which doesn't use any special functions, only the well known "classic" series.

$$I=\int_0^1 \frac{\ln{(x^2+1)}}{x+1}dx=\sum_{k,n=0}^\infty \frac{(-1)^{k+n}}{k+1} \int_0^1 x^{2(k+1)+n}dx=\sum_{k,n=0}^\infty \frac{(-1)^{k+n}}{(k+1)(2k+n+3)}$$

Let's use partial fractions:

$$\frac{1}{(k+1)(2k+n+3)}=\frac{a}{k+1}+\frac{b}{2k+n+3} \\ 2a+b=0 \\ (n+3)a+b=1 \\ (n+1)a=1 \\ a=\frac{1}{n+1} \\ b=-\frac{2}{n+1}$$

So we have:

$$I=\sum_{k,n=0}^\infty \frac{(-1)^n}{n+1} \frac{(-1)^k}{k+1}-2 \sum_{k,n=0}^\infty \frac{(-1)^n}{n+1} \frac{(-1)^k}{2k+n+3}=\ln^2 2-2\sum_{n=0}^\infty \frac{(-1)^n}{n+1} \sum_{k=0}^\infty \frac{(-1)^k}{2k+n+3}$$

The second series can be represented the following way, separating even and odd terms:

$$\sum_{n=0}^\infty \frac{(-1)^n}{n+1} \sum_{k=0}^\infty \frac{(-1)^k}{2k+n+3}=\sum_{l=0}^\infty \frac{1}{2l+1} \sum_{k=0}^\infty \frac{(-1)^k}{2(k+l)+3}-\sum_{l=0}^\infty \frac{1}{2l+2} \sum_{k=0}^\infty \frac{(-1)^k}{2(k+l)+4}$$

Now let's deal with the inner series separately:

1) $$\sum_{k=0}^\infty \frac{(-1)^k}{2(k+l)+3}=(-1)^{l+1} \sum_{k=l+1}^\infty \frac{(-1)^k}{2k+1}=(-1)^{l+1} \left( \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}- \sum_{k=0}^l \frac{(-1)^k}{2k+1} \right)= \\ =(-1)^{l+1} \left( \frac{\pi}{4}- \sum_{k=0}^l \frac{(-1)^k}{2k+1} \right)$$

The first term is the well known Leibniz series for $\pi$.

2) $$\sum_{k=0}^\infty \frac{(-1)^k}{2(k+l)+4}=\frac{(-1)^{l+1}}{2} \sum_{k=l+1}^\infty \frac{(-1)^k}{k+1}=\frac{(-1)^{l+1}}{2} \left( \sum_{k=0}^\infty \frac{(-1)^k}{k+1}- \sum_{k=0}^{l} \frac{(-1)^k}{k+1} \right)= \\ =\frac{(-1)^{l+1}}{2} \left(\ln 2- \sum_{k=0}^{l} \frac{(-1)^k}{k+1} \right)$$

The first terms of each can now be summed w.r.t. $l$:

$$\sum_{l=0}^\infty \frac{1}{2l+1} (-1)^{l+1} \frac{\pi}{4}=-\frac{\pi^2}{16}$$

$$\frac{1}{2} \sum_{l=0}^\infty \frac{1}{l+1} \frac{(-1)^{l+1}}{2} \ln 2=-\frac{\ln^2 2}{4}$$

At this point we have:

$$I=\frac{\ln^2 2}{2}+\frac{\pi^2}{8}-\sum_{l=0}^\infty (-1)^l \left(\frac{2}{2l+1} \sum_{k=0}^l \frac{(-1)^k}{2k+1}-\frac{1}{2}\frac{1}{l+1} \sum_{k=0}^{l} \frac{(-1)^k}{k+1} \right)$$

Well, I tried. Now it seems that there's at least two similar questions with some good answers, but I'll still leave this attempt here in case it could be finished.

We need to prove that:

$$\sum_{l=0}^\infty (-1)^l \left(\frac{2}{2l+1} \sum_{k=0}^l \frac{(-1)^k}{2k+1}-\frac{1}{2}\frac{1}{l+1} \sum_{k=0}^{l} \frac{(-1)^k}{k+1} \right)=\frac{7 \pi^2}{48}-\frac{\ln^2 2}{4}$$

which seems like a long shot, even though numerically it's true.

Mostly I just wanted to show how the squared logarithm and $\pi$ appear in the answer.