Integral $\int_0^1 \frac{\sqrt[4]{\ln(1+t)} +\sqrt {\ln(1+t)}}{t^2+1} dt$

Question

Consider $$ I = \int_0^1 \frac{\sqrt[4]{\ln(1+t)} +\sqrt {\ln(1+t)}}{t^2+1} dt$$ Find a closed form for $I$ and show that $I > 1$.

We know that

$$\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx = \frac{\ln(2) \pi}{8}$$

For instance by these $2$ proofs :

Consider:

$$I(a) = \int_0^1 \frac{\ln (1+ax)}{1+x^2} \, dx$$ then the derivative $I'$ is: $$I'(a) = \int_0^1 \frac{x}{(1+ax)(1+x^2)} \, dx = \frac{2 a \arctan x - 2\ln (1+a x) + \ln (1+x^2)}{2(1+a^2)} \Big|_0^1\\ = \frac{\pi a + 2 \ln 2}{4(1+a^2)} - \frac{\ln (1+a)}{1+a^2}$$ Hence: $$I(1) = \int_0^1 \left( \frac{\pi a + 2 \ln 2}{4(1+a^2)} - \frac{\ln (1+a)}{1+a^2} \right) \, da \\ 2 I(1) = \int_0^1 \frac{\pi a + 2 \ln 2}{4(1+a^2)} \, da = \frac{\pi}{4} \ln 2$$ Divide both sides by $2$ and you're done.

Or this proof

Put $x = \tan\theta$, then your integral transforms to $$I= \int_{0}^{\pi/4} \ln(1+\tan\theta) \ d\theta. \tag{1}$$

Now using the property that $$\int_{0}^{a} f(x) \ dx = \int_{0}^{a} f(a-x) \ dx,$$ we have $$I = \int_{0}^{\pi/4} \ln\biggl(1+\tan\Bigl(\frac{\pi}{4}-\theta\Bigr)\biggr) \ d\theta = \int_{0}^{\pi/4} \ln\biggl(\frac{2}{1+\tan\theta} \biggr) \ d\theta.\tag{2}$$

Adding $(1)$ and $(2)$ we get $$2I = \int_{0}^{\pi/4} \ln(2) \ d\theta\Rightarrow I= \ln(2) \cdot \frac{\pi}{8}.$$

Or this proof

$$\int_0^1\frac{\ln (x+1)}{x^2+1}dx=\int_1^0\frac{\ln\frac{2}{1+t}}{\left(\frac{1-t}{1+t}\right)^2+1}\cdot\left\{-\frac{2}{(1+t)^2}\right\}dt =\int_0^1\frac{\ln 2-\ln (1+t)}{t^2+1}\ dt.$$

and with some trivial algebra you get the solution.

Similarly we can compute the easier integral :

$$\int_0^1 \frac{\sqrt {\ln(1 + t)}}{t + 1} dt = 2/3 \ln^{3/2}(2)$$

or this one

$$\int_0^1\frac{\sqrt{\ln(1+t)}}{\sqrt{t+1}}dt=2\sqrt{\ln(4)}-\sqrt{2\pi}\operatorname{erfi}\Big(\sqrt{\ln(2)/2}\Big)$$

and many similar ones.

We also have

$$\int\sqrt{\ln(1+x)}dx=(x+1)\sqrt{\ln(1+x)}-1/2\sqrt\pi\operatorname{erfi}\Big(\sqrt{\ln(1+x)}\Big)+C$$

$$\int\frac{dx}{\sqrt{\ln(1+x)}}=\sqrt\pi\operatorname{erfi}\Big(\sqrt {\ln(1+x)}\Big)+C$$

and there is also a long closed form for

$$\int \ln(1+x) \sqrt {1+x^2} dx $$

or a slighly shorter closed form for

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

I am not sure that the indefinite $\int \frac{\sqrt[4]{\ln(1+t)} +\sqrt {\ln(1+t)}}{t^2+1} dt$ can be solved by standard functions.

But the definite integral with range $\int_0^1$ seems more likely to have a closed form.

Maybe some contour integral is helpful.

Or a series expansion for say $\sqrt{\ln(1+x)}$. Lagrange inversion might help. Or the more general Lagrange–Bürmann formula:

see

https://en.wikipedia.org/wiki/Lagrange_inversion_theorem

Faa di Bruno might also help

see

https://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula

see also the related Gregory coefficients :

https://en.wikipedia.org/wiki/Gregory_coefficients

In fact $\frac1{\ln(1+x)}$ has been studied alot, and even its generalization $\ln(1+x)^t$ but I was not able to find free sources. They are often called "generalized" Gregory or Bernoulli coefficients and they satisfy many properties.

Answer 1

We want to evaluate $$ I \;=\; \int_{0}^{1}\frac{\sqrt[4]{\log(1+t)}+\sqrt{\log(1+t)}}{1+t^{2}}\;\text{d}t \;=\; I_{1/4}+I_{1/2}, $$ where $$ I_\nu \;:=\; \int_{0}^{1}\frac{\log^\nu(1+t)}{1+t^{2}}\;\text{d}t, \qquad \nu>-1. $$

Put $t=u-1$ ($\text{d}t=\text{d}u$). Then $u\in[1,2]$ and $$ I_\nu \;=\; \int_{1}^{2}\frac{\log^\nu(u)}{(u-1)^{2}+1}\;\text{d}u. $$ Since $$ (u-1)^{2}+1=(u-1+i)(u-1-i) \;=\;\frac{1}{2i}\Bigl[\frac{1}{u-(1-i)}-\frac{1}{u-(1+i)}\Bigr], $$ we split the integral as $$ I_\nu=\frac{1}{2i}\bigl(J_\nu^{+}-J_\nu^{-}\bigr),\qquad J_\nu^{\pm}:=\int_{1}^{2}\frac{\log^\nu(u)}{u-(1\mp i)}\;du. $$

Set $u=e^{t}$ ($\text{d}u=e^{t}\,\text{d}t$). Then $t\in[0,\log(2)]$ and \begin{align*} J_\nu^{\pm} &=\int_{0}^{\log(2)}\frac{t^{\nu}\,e^{t}}{e^{t}-(1\mp i)}\,\text{d}t =\int_{0}^{\log(2)}t^{\nu}\,\text{d}t \;+\;(1\mp i)\int_{0}^{\log(2)}\frac{t^{\nu}}{e^{t}-(1\pm i)}\,\text{d}t. \end{align*}

The first term is the same in $J_\nu^{-}$ and $J_\nu^{+}$, so it vanishes in $J_\nu^{-}-J_\nu^{+}$. Hence $$ I_\nu =\operatorname{Im}\!\Bigl[ (1-i)\int_{0}^{\log(2)}\frac{t^{\nu}}{e^{t}-(1-i)}\,\text{d}t - (1+i)\int_{0}^{\log(2)}\frac{t^{\nu}}{e^{t}-(1+i)}\,\text{d}t \Bigr]. $$ This last integral can be expressed in terms of the incomplete gamma $\gamma(s,x)$ cannot be reduced to elementary functions, so it is unavoidable in any closed form.