Evaluate the integral from 0 to 1 of ln(x) divided by (x to the 0.9 times (1 − x) to the 0.1), with respect to x.

Step-by-step solution

1. 1

   Put the integrand in Beta-function shape. Rewrite it as

   $I = \int_0^1 x^{-0.9}(1-x)^{-0.1}\ln x\,dx$

   and compare with

   $B(a,b) = \int_0^1 x^{a-1}(1-x)^{b-1}dx$

   Matching exponents gives $a - 1 = -0.9$ so $a = 0.1$, and $b - 1 = -0.1$ so $b = 0.9$. Both exponents exceed $-1$, so the integral converges at each endpoint despite the singularities.

2. 2

   Produce the logarithm by differentiating in a. Differentiating under the integral sign brings down exactly $\ln x$:

   $\frac{\partial}{\partial a}B(a,b) = \int_0^1 x^{a-1}(1-x)^{b-1}\ln x\,dx$

   so $I = \left.\partial_a B(a,b)\right|_{a=0.1,\,b=0.9}$.

3. 3

   Differentiate correctly — the trap is the third gamma. From $B(a,b) = \dfrac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$, logarithmic differentiation with $b$ held fixed gives

   $\frac{\partial}{\partial a}B(a,b) = B(a,b)\bigl[\psi(a) - \psi(a+b)\bigr]$

   It is tempting to note $a + b = 1$ at the evaluation point and write $B = \Gamma(a)\Gamma(b)$ before differentiating — but $a+b$ is not constant as $a$ varies, so dropping $\Gamma(a+b)$ loses the entire $-\psi(a+b)$ term.

4. 4

   Evaluate the Beta value with the reflection formula. At $a = 0.1$, $b = 0.9$ we do have $a + b = 1$, so $\Gamma(a+b) = \Gamma(1) = 1$ and

   $B(0.1, 0.9) = \Gamma(0.1)\Gamma(0.9) = \frac{\pi}{\sin(0.1\pi)} = \frac{\pi}{\sin(\pi/10)} \approx 10.16640$

5. 5

   Assemble the digamma factor. With $\psi(1) = -\gamma$ (Euler-Mascheroni, $\gamma \approx 0.5772157$) and $\psi(0.1) \approx -10.4237549$:

   $\psi(0.1) - \psi(1) = -10.4237549 + 0.5772157 = -9.8465392$

   so

   $I = \frac{\pi}{\sin(\pi/10)}\bigl(\psi(0.1) + \gamma\bigr)$

6. 6

   Compute and verify numerically.

   $I \approx 10.1664074 \times (-9.8465392) \approx -100.10393$

   Independent numerical integration — substituting $x = u^{10}$ near $0$ and $x = 1 - v^{10}$ near $1$ to tame both singularities, then applying $300$-node Gauss-Legendre — gives $-100.10393$, confirming the closed form. (Omitting the $-\psi(a+b)$ term would have given $-105.97$, about $6\%$ too negative.) The result is negative because $\ln x < 0$ on the whole interval.