Integral of the power of Euclidean norm over a line segment in $R^n$

Question

Let $x,y \in \mathbb{R}^n$, $p>0$, and $||\cdot||_2$ the Euclidean norm.

What is

$$ \int_0^1 ||x + s(y-x)||_2^p \; ds?$$

If it makes things easier, one can also assume $p>1$ or $p=1$ for a start.

Clearly, this is a line integral of the line connecting $x$ and $y$ and it seems to me that this should be easy but I cannot quite get it done.

The only special case I found that can be easily solved is if $x$ and $y$ lie on a line through the origin, i.e., if $y = \alpha x$ with $\alpha>1$. In this case, we have

$$ \int_0^1 ||x + s(\alpha x-x)||_2^p \; ds = \int_0^1 || x (1 + s(\alpha-1))||_2^p \; ds = || x ||_2^p \; \int_0^1 (1 + s(\alpha-1))^p \; ds = || x ||_2^p \frac{\alpha^{p+1}-1}{(p+1)(\alpha-1)}. $$

Answer 1

The integral takes the form $$I=\int_0^1\big(as^2+2bs+c)^{p/2}\,ds$$ where $a=\|y-x\|^2,\ b=\langle x,y-x\rangle,\ c=\|x\|^2$ with the Euclidean norm $\|\cdot\|$.

• If $p\in2\mathbb N$, the integrand is a polynomial in $s$, which is easy to calculate.
• If $p\in2\mathbb N+1$, rewrite the integrand into $\alpha\big(1+\beta(s+\gamma)^2\big)^{p/2}$ for some $\alpha,\beta>0,\gamma\in\mathbb R$ and use the substitution $\tan(u)=\sqrt\beta(s+\gamma)$, it follows $$I=\frac\alpha{\sqrt\beta}\int_{\tan^{-1}(\sqrt\beta\gamma)}^{\tan^{-1}(\sqrt\beta(1+\gamma))}\sec^{p+2}(u)\,du.$$ The integral can then be tackled using reduction formula.
• If $p\notin\mathbb N$, it is impossible to find a closed-form solution in general, so numerical integration might be your friend.

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As we can see, the integral is pretty tedious to tackle when $a,b,c,p$ are general. This is because although the integral is taken on a simple line segment, the function being integrated may not be simple (if $p\notin\mathbb N$), therefore the resulting integral could be hard to deal with.

However, there are some special cases:

• When $b^2=ac$, the integrand is simplified to $a^{p/2}(s+b/a)^p$, which is easily solvable. But note that $b^2=ac\ \Leftrightarrow\ \|x\|\|y\|=|\langle x,y\rangle|\ \Leftrightarrow\ x$ and $y$ lie on the same line (already shown in the question).

• When $p=2$, the integral has a simplified value $I=\frac13\big(\|x+y\|^2-\langle x,y\rangle\big)$.

• When the Euclidean norm $\|\cdot\|$ is replaced by $\|\cdot\|_{L_t}$ with $t\in\{1,p,\infty\}$, the integral has a closed-form value in terms of the coordinates of $x$ and $y$.