4

One can find the antiderivate with help of the partial fraction method introducing complex numbers: $$\frac{x\ln(x+1)}{x^2+1}=\frac{1}{2}\left(\frac{\ln(x+1)}{x-i}+\frac{\ln(x+1)}{x+i}\right).$$ The result is a complicated function containing dilogarithms and logarithms with complex arguments. The value of the definite integral turns out to be $$\int_0^1 \frac{x\ln(x+1)}{x^2+1} dx = \frac{1}{96}(\pi^2+12\ln^2(2)).$$

Is there a way to find that value more directly - without introducing complex numbers?

Ben Hur
  • 151

1 Answers1

2

Rewrite the integral as

$$I=\int_0^1 \frac{x\log (x+1)}{x^2+1}dx =\int_0^1\int_0^1\frac{x^2\:dydx}{(yx+1)(x^2+1)}$$

A partial fraction and integration order reverse later we get

$$\int_0^1\int_0^1 \frac{xy}{(y^2+1)(x^2+1)}-\frac{1}{(y^2+1)(x^2+1)} +\frac{1}{xy+1}-\frac{y^2}{(y^2+1)(xy+1)}\:dxdy = \int_0^1\frac{y}{2(y^2+1)}\log(2) -\frac{\pi}{4}\frac{1}{y^2+1}+ \frac{\log(1+y)}{y}-\frac{y\log(y+1)}{y^2+1}\:dy$$

The first two terms simplify easily and the last term is the integral we started with. The third term is given by Taylor series

$$\sum_{n=1}^\infty \frac{(-1)^n}{n}\int_0^1x^{n-1}dx = \sum_{n=1}^\infty \frac{(-1)^n}{n^2} =\frac{\pi^2}{12}$$

thus the equation simplifies to $$\implies I = \frac{\log^22}{4}-\frac{\pi^2}{16}+\frac{\pi^2}{12}- I$$

therefore $I = \frac{1}{96}(\pi^2 +12\log^22)$

Ninad Munshi
  • 34,407
  • 2
    The partial fraction term is somehow magical. Why do you add the factor $y^2+1$ in the denominator? – Ben Hur Feb 22 '21 at 10:33
  • Does the method work for the generalisation $\int_0^1\frac{x\ln(x+1)}{ax^2+1}dx$ ? – Ben Hur Feb 22 '21 at 11:10
  • @BenHur sure, I don't see why it wouldn't – Ninad Munshi Feb 22 '21 at 11:35
  • @BenHur the beauty of partial fractions with multiple letters is that there isn't any uniqueness because you could pretend one letter is "the variable", the other, or even both! The original expression, for example was already fully partial fractional for $y$ – Ninad Munshi Feb 22 '21 at 11:37
  • I tried hard to apply your ideas to $\int_0^1 \frac{x\ln(x+1)}{x^2+3}$ but failed. Any help? – Ben Hur Mar 08 '21 at 16:52