How to Compute ∫₀¹ xᵐ(1 − xⁿ)ᵖ dx

For real parameters, the standard interpretation is that $n>0$ and the integral is an improper integral at the endpoints if needed. Under that interpretation, its value is a Beta function. The formula also gives a direct test for when the integral converges.

The Beta-function formula

For $n>0$, $m>-1$, and $p>-1$,

$$\int_0^1 x^m(1-x^n)^p\,dx =\frac{1}{n}\operatorname B\left(\frac{m+1}{n},p+1\right) =\frac{1}{n}\frac{\Gamma\left(\frac{m+1}{n}\right)\Gamma(p+1)}{\Gamma\left(\frac{m+1}{n}+p+1\right)}.$$

Here the Beta function is defined, for $a,b>0$, by

$$\operatorname B(a,b)=\int_0^1 t^{a-1}(1-t)^{b-1}\,dt.$$

The Gamma-function form follows from $\operatorname B(a,b)=\Gamma(a)\Gamma(b)/\Gamma(a+b)$.

Derivation by substitution

Set

$$t=x^n.$$

Since $n>0$, this substitution maps $x\in[0,1]$ to $t\in[0,1]$. Solving for $x$ and differentiating gives

$$x=t^{1/n},\qquad dx=\frac{1}{n}t^{1/n-1}\,dt.$$

Also, $x^m=t^{m/n}$ and $1-x^n=1-t$. Therefore,

$$\begin{aligned} \int_0^1 x^m(1-x^n)^p\,dx &=\frac{1}{n}\int_0^1 t^{m/n+1/n-1}(1-t)^p\,dt\\ &=\frac{1}{n}\int_0^1 t^{(m+1)/n-1}(1-t)^{(p+1)-1}\,dt. \end{aligned}$$

This is the defining Beta integral with $a=(m+1)/n$ and $b=p+1$.

When does the integral converge?

The endpoint conditions come from the behavior of the integrand near $0$ and $1$:

• As $x\to0^+$, $(1-x^n)^p\to1$, so the integrand behaves like $x^m$. Its integral near zero is finite exactly when $m>-1$.
• As $x\to1^-$, $1-x^n$ behaves like $n(1-x)$. Thus the integrand behaves like a constant times $(1-x)^p$, which is integrable near one exactly when $p>-1$.

These are also the conditions that make both Beta-function arguments positive. If either condition fails, the improper integral diverges. This convergence statement assumes real $m,p$ and $n>0$; other choices, especially $n<0$, change the endpoint behavior and require a separate analysis.

Worked example

Evaluate

$$\int_0^1 (1-x^2)^{1/2}\,dx.$$

This is the case $m=0$, $n=2$, and $p=1/2$, which satisfies the convergence conditions. The formula gives

$$\begin{aligned} \int_0^1 (1-x^2)^{1/2}\,dx &=\frac12\operatorname B\left(\frac12,\frac32\right)\\ &=\frac12\frac{\Gamma(1/2)\Gamma(3/2)}{\Gamma(2)}\\ &=\frac12\frac{\sqrt{\pi}(\sqrt{\pi}/2)}{1} =\frac{\pi}{4}. \end{aligned}$$

The result also has a geometric check: the graph $y=\sqrt{1-x^2}$ on $0\le x\le1$ is a quarter of the unit circle, whose area is $\pi/4$.

Common mistakes

• Dropping the factor $1/n$. It comes from differentiating $x=t^{1/n}$ and is essential.
• Using the wrong Beta arguments. The power of $t$ is $(m+1)/n-1$, so the first argument is $(m+1)/n$, not $m/n$.
• Forgetting the added one in the second argument. The factor $(1-t)^p$ corresponds to $b-1=p$, so $b=p+1$.
• Applying the formula outside its convergence range. For real parameters and $n>0$, require both $m>-1$ and $p>-1$ before treating the result as a finite integral.

For a particular choice of $m,n,p$, first check those conditions, then substitute into the Beta formula. If the arguments are familiar, Gamma identities such as $\Gamma(z+1)=z\Gamma(z)$ can often simplify the answer.