Solving double integral $\int _0^1\int _1^x\frac{\ln\left(1+t\right)}{t \ \sqrt{x}} \ \mathrm dt \ \mathrm dx$

Question

How to solve the following integral? $$\int _0^1\int _1^x\frac{\ln\left(1+t\right)}{t \sqrt{x}} \ \mathrm dt \ \mathrm dx$$

I tried to change the variables by using subsitution $(1+t) = u$ and $t \sqrt{x} = v$ but it didn't simplify the integral.

Any hints?

Answer 1

From Wolfram Alpha $\int _1^x\frac{\ln(1+t)}{t} \ dt=-Li_2(-x)-\frac{\pi^2}{12}$, so it doesn't have a 'regular' function as an aswer.

Also the final answer is $-1.055$

Answer 2

As noted before the inner integral is not doable using a suitable substitution. Anyway, this is not a real problem by accepting the usage of special functions, in particular by invoking the Dilogarithm Function. Doing so one can get back to elementary integrals yielding the result.

We shall proceed in the following manner \begin{align*} \int_0^1\int_1^x\frac{\log(1+t)}t{\rm d}t\frac{{\rm d}x}{\sqrt x}&=\int_0^1[-{\rm Li}_2(-t)]_1^x\frac{{\rm d}x}{\sqrt x}\\ &=\int_0^1\left[-{\rm Li}_2(-x)-\frac{\pi^2}{12}\right]\frac{{\rm d}x}{\sqrt x}\\ &=\left[2\sqrt x\left(-{\rm Li}_2(-x)-\frac{\pi^2}{12}\right)\right]_0^1-2\int_0^1\frac{\log(1+x)}{\sqrt x}{\rm d}x\\ &=-4\int_0^1\log(1+x^2){\rm d}x\\ &=-4[x\log(1+x^2)]_0^1+8\int_0^1\frac{x^2}{1+x^2}{\rm d}x\\ &=-4\log(2)+8\int_0^11-\frac1{1+x^2}{\rm d}x\\ &=-4\log(2)+8-8[\arctan(x)]_0^1 \end{align*}

$$\therefore~\int_0^1\int_1^x\frac{\log(1+t)}t{\rm d}t\frac{{\rm d}x}{\sqrt x}~=~8-2\pi-4\log(2)$$

Note, however, that the Dilogarithm is only used in between to make some calculations look prettier and is almsot immediately removed again by via IBP. Therefore, it might be possible to dodge the utilisation of the Dilogarithm by simply denoting the intermediate anti-derivative as, lets say, $F(t)$ and observe that the value of $F(t)$ at $1$ is $0$ and at $0$ is defined (as $\sqrt x$ is $0$ at this point). The decision is up to you.

Answer 3

Start with changing the order of the integration

$$\int_0^1\int_1^x\frac{\ln(1+t)}{t\sqrt{x}}dtdx=\int_0^1\int_t^0\frac{\ln(1+t)}{t\sqrt{x}}dxdt=-2\int_0^1\frac{\ln(1+t)}{\sqrt{t}}dt$$

$$\overset{\sqrt{t}=u}{=}-4\int_0^1\ln(1+u^2)du\overset{IBP}{=}-4\ln2+8\int_0^1\frac{u^2}{1+u^2}du$$

$$=-4\ln2+8\left(1-\frac{\pi}{4}\right)=\boxed{8-2\pi-4\ln2}$$

and this way we avoided using any special function.