How can I prove $\int_{0}^{1} \frac {x-1}{\log(x) (1+x^3)}dx=\frac {\log3}{2}$

Note that $\int_0^1 x^s\,ds=\frac{x-1}{\log(x)}$. Then, we have

$$\int_0^1\frac{x-1}{\log(x)(x^3+1)}\,dx=\int_0^1\int_0^1 \frac{x^s}{(x^3+1)}\,ds\,dx$$

Now we apply Fubini's Theorem to interchange the order of integration to reveal

$$\int_0^1\frac{x-1}{\log(x)(x^3+1)}\,dx=\int_0^1\int_0^1 \frac{x^s}{(x^3+1)}\,dx\,ds$$

Next, we expand the denominator in a geometric series to find that

$$\begin{align} \int_0^1\frac{x-1}{\log(x)(x^3+1)}\,dx&=\sum_{n=0}^\infty (-1)^n\int_0^1\int_0^1 x^{s+3n}\,dx\,ds\\\\ &=\sum_{n=0}^\infty (-1)^n \log\left(\frac{3n+2}{3n+1}\right) \end{align}$$

Can you finish now?

BONUS:

To evaluate the final series we appeal to the digamma function, its relationship with the Gamma function, and Euler's reflection formula. Proceeding, we write

$$\begin{align} \sum_{n=0}^\infty (-1)^n\log\left(\frac{3n+2}{3n+1}\right)&=\int_0^1 \sum_{n=0}^\infty (-1)^n \frac1{s+3n+1}\,ds\\\\ &=\int_0^1 \sum_{n=0}^\infty\left(\frac1{6n+s+1}-\frac1{6n+s+4}\right)\,ds\\\\ &=\frac16\int_0^1\left(\psi((s+4)/6)-\psi((s+1)/6)\right)\,ds\\\\ &=\log\left(\frac{\Gamma(5/6)\Gamma(1/6)}{\Gamma(2/3)\Gamma(1/3)}\right)\\\\ &=\log\left(\frac{\sin(2\pi/3)}{\sin(5\pi/6)}\right)\\\\ &=\log(\sqrt 3) \end{align}$$

as expected!


Note

$$I=\int_{0}^{1} \frac {x-1}{\ln x (1+x^3)}dx \overset{x\to\frac1x}= \frac12\int_{0}^{\infty} \frac {x-1}{\ln x (1+x^3)}dx$$

Let $J(a) = \int_{0}^{\infty} \frac {x^a-1}{\ln x (1+x^3)}dx$. Then $J’(a) = \int_{0}^{\infty} \frac {x^a}{1+x^3}dx=\frac\pi3\csc\frac{\pi(a+1)}3 $. Thus,

$$I=\frac12 J(1) =\frac12\int_0^1J’(a)da=\frac\pi6\int_0^1\csc\frac{\pi(a+1)}3da=\frac{\ln3}2 $$


Alternatively to Mark Viola's approach, use the geometric series to see $$\small\int_0^1\frac{x-1}{x^3+1}\frac{{\rm d}x}{\log x}=\sum_{n\ge0}(-1)^n\int_0^1\frac{x^{3n+1}-x^{3n}}{\log x}\,{\rm d}x=\sum_{n\ge0}(-1)^{n+1}\int_0^\infty\frac{e^{-(3n+2)x}-e^{-(3n+1)x}}x\,{\rm d}x$$ The latter is a Frullani integral and evaluates as $$\int_0^\infty\frac{e^{-(3n+2)x}-e^{-(3n+1)x}}x\,{\rm d}x=-\log\left(\frac{3n+2}{3n+1}\right)$$ and thus arriving at $$\int_0^1\frac{x-1}{x^3+1}\frac{{\rm d}x}{\log x}=\sum_{n\ge0}(-1)^n\log\left(\frac{3n+2}{3n+1}\right)$$ aswell.

Tags:

Integration

Calculus

Definite Integrals

Related

Does a function $f$ with the following property exist? How to evaluate $\int_0^{\pi/2} x\ln^2(\sin x)\textrm{d}x$ in a different way? If two zero sets are homeomorphic, is the ring of polynomials over the sets also homeomorphic? A sum of series with the inverse squared central binomial coefficient Number of permutations of $D,D,D,O,O,O,G,G,G$ such that no two $D$ are adjacent and no two $G$ are adjacent Can a sum of $n$ squares be expressed as the sum of $n/2$ squares? For a positive integer $n\geq 2$ with divisors $1=d_1<d_2<\cdots<d_k=n$, prove that $d_1d_2+d_2d_3+\cdots+d_{k-1}d_k<n^2$ Suppose $\angle BAC = 60^\circ$ and $\angle ABC = 20^\circ$. A point $E$ inside $ABC$ satisfies $\angle EAB=20^\circ$ and $\angle ECB=30^\circ$. Does a nontrivial finite solvable group have a subgroup of prime power index for each prime divisor? If $V_n(a)$ counts sign changes in the sequence $\cos a, \cos2a,\cos3a,\ldots,\cos na,$ show that $\lim_{n\to\infty}\frac{V_n(a)}n=\frac{a}\pi$ probability group students to maximize expectation Is there an efficient way of showing $\int_{-1}^{1} \ln\left(\frac{2(1+\sqrt{1-x^2})}{1+x^2}\right)dx = 2$?

Recent Posts