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Compute $$I = \int_{0}^{1} \int_0^1 \frac{1}{1+y(x^2-x)}dydx$$

Here are my steps: $$\begin{split} I &=\int_{0}^{1} \left(\int_0^1 \frac{dy}{1+y(x^2-x)}\right)dx\\ &=\int_{0}^{1} \left[\frac{\ln(1+y(x^2-x))}{x^2-x}\right]_0^1dx\\ &=\int_{0}^{1} \left[\frac{\ln(1+(1)(x^2-x))}{x^2-x} -\frac{\ln(1+(0)(x^2-x))}{x^2-x}\right]dx\\ &=\int_{0}^{1} \frac{\ln(1+x^2-x)}{x^2-x}dx \end{split} $$ And here I can't find any substitution to solve this integral.

Can anyone help me?

By the way, I also used Simpson's 3/8 method to find the approximation and got $1.063$.

But I want to find it using Calculus.

Zacky
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Pankaj Solanki
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3 Answers3

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Switching the order of integration, we get $$4 \int_{y=0}^1 \frac{1}{\sqrt{y(4-y)}} \tan^{-1} \sqrt{\frac{y}{4-y}}\, dy.$$ Then the substitution $$u = \tan^{-1} \sqrt{y/(4-y)}, \quad du = \frac{1}{1+\frac{y}{4-y}} \cdot \frac{2}{(4-y) \sqrt{y(4-y)}} \, dy = \frac{1}{2 \sqrt{y (4-y)}} \, dy,$$ gives $$4 \left[\left(\tan^{-1} \sqrt{\frac{y}{4-y}} \right)^2\right]_{y=0}^1 = \frac{\pi^2}{9}.$$

heropup
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$$I=\int_0^1 \frac{\ln(1-x+x^2)}{x(x-1)}dx=-\int_0^1 \frac{\ln(1-x(1-x))}{x}dx-\int_0^1\frac{\ln(1-x(x-1))}{1-x}dx$$ The substitution $1-x\to x $ in the second integral reveals that: $$\int_0^1\frac{\ln(1-x(1-x))}{1-x}dx=\int_0^1\frac{\ln(1-x(1-x))}{x}dx$$ $$\Rightarrow I=-2\int_0^1\frac{\ln(1-x(1-x))}{x}dx=2\int_0^1\frac{\ln(1+x)-\ln(1+x^3)}{x}dx$$ But we are lucky again, because: $$\int_0^1\frac{\ln(1+x^3)}{x}dx\overset{x=t^{1/3}}=\frac13\int_0^1 \frac{\ln(1+t)}{t^{1/3}}\,t^{1/3-1}dt\overset{t=x}=\frac13\int_0^1\frac{\ln(1+x)}{x}dx$$ Also we can use this equality which will lead to: $$ I=\frac43 \int_0^1 \frac{\ln(1+x)}{x}dx=-\frac23 \int_0^1 \frac{\ln x}{1-x}dx=-\frac23 \sum_{n=0}^\infty \int_0^1 x^n\ln x dx=\frac23 \sum_{n=1}^\infty \frac{1}{n^2}$$ Combining with heropup's answer we can deduce that $$\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$$ I guess we can add one more to this big list Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$ (Basel problem).

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Zacky
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1

Here is an alternate way that avoids your integral.

Writing the integrand in terms of a geometric sum we have: $$\frac{1}{1 + y(x^2 - x)} = \sum_{n = 0}^\infty y^n (x - x^2)^n, \quad |x|, |y| < 1,$$ then \begin{align} I &= \int_0^1 \int_0^1 \frac{1}{1 + y(x^2 - x)} \, dy dx\\ &= \sum_{n = 0}^\infty \int_0^1 y^n \, dy \int_0^1 x^n (1 - x)^n \, dx\\ &= \sum_{n = 0}^\infty \frac{1}{n + 1} \cdot \operatorname{B}(n + 1, n + 1)\\ &= \sum_{n = 0}^\infty \frac{1}{n + 1} \cdot \frac{(n!)^2}{(2n + 1)!}\\ &= \sum_{n = 0}^\infty \frac{1}{(n + 1)(2n + 1) \binom{2n}{n}}.\tag1 \end{align} Here $\operatorname{B}(x,x)$ denoted the Beta function while $\binom{2n}{n} = \frac{(2n)!}{(n!)^2}$ is the central binomial coefficient.

To find the sum in (1), the following (reasonably?) well known Maclaurin series can be recalled (for a proof, see here) $$\frac{\sin^{-1} x}{\sqrt{1 - x^2}} = \sum_{n = 0}^\infty \frac{2^{2n} x^{2n + 1}}{(2n + 1) \binom{2n}{n}}, \quad |x| < 1.$$ Enforcing a substitution of $x \mapsto \sqrt{x}$, after rearranging one has $$\frac{\sin^{-1} \sqrt{x}}{\sqrt{x} \sqrt{1 - x}} = \sum_{n = 0}^\infty \frac{2^{2n} x^n}{(2n + 1) \binom{2n}{n}}, \quad |x| < 1.$$ Integrating both sides with respect to $x$ from $0$ to $\frac{1}{4}$ leads to $$\sum_{n = 0}^\infty \frac{1}{(n + 1)(2n + 1) \binom{2n}{n}} = 4 \int_0^{\frac{1}{4}} \frac{\sin^{-1} \sqrt{x}}{\sqrt{x} \sqrt{1 - x}} \, dx.$$ Making a substitution of $x = \sin^2 u$ in the integral readily leads to \begin{align} \sum_{n = 0}^\infty \frac{1}{(n + 1)(2n + 1) \binom{2n}{n}} &= 8 \int^{\frac{\pi}{6}}_0 u \, du = \frac{\pi^2}{9}, \end{align} from which it follows that: $$\int_0^1 \int_0^1 \frac{1}{1 + y(x^2 - x)} \, dy dx = \frac{\pi^2}{9}.$$

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omegadot
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