Home » Blog » Prove a formula for the integral from 0 to x of tm(1+t)n

Prove a formula for the integral from 0 to x of tm(1+t)n

by RoRi

Define a function

    \[ F(m,n) = \int_0^x t^m (1+t)^n \, dt \qquad m > 0, \quad n > 0. \]

Prove that the following relation holds,

    \[ (m+1)F(m,n) + nF(m+1,n-1) = x^{m+1}(1+x)^n. \]

Using this find the value of F(10,2).

First, we evaluate F(m,n) by parts, letting

    \begin{align*} u &= (1+t)^n & du &= n (1+t)^{n-1} \\[9pt] dv &= t^m \, dt & v &= \frac{1}{m+1} t^{m+1}. \end{align*}

This gives us

    \begin{align*} F(m,n) &= \int_0^x t^m (1+t)^n \, dt \\[9pt] &= \left( uv - \int v \, du \right) \Bigr \rvert_0^x \\[9pt] &= \frac{t^{m+1}}{m+1}(1+t)^n \Bigr \rvert_0^x - \frac{n}{m+1} \int_0^x t^{m+1}(1+t)^{n-1} \, dt \\[9pt] &= \frac{x^{m+1}}{m+1} (1+x)^n - \frac{n}{m+1} F(m+1,n-1). \end{align*}

Multiplying both sides by m+1 we have

    \begin{align*} (m+1) F(m,n) &= x^{m+1}(1+x)^n - nF(m+1,n-1) \\[9pt] \implies (m+1)F(m,n) + nF(m+1,n-1) &= x^{m+1}(1+x)^n. \qquad \blacksquare \end{align*}

Then, to compute F(10,2) we first compute F(11,1) to use in our recurrence,

    \begin{align*} F(11,1) = \int_0^x t^{11} (1+t) \, dt &= \int_0^x (t^{11} + t^{12} ) \, dt \\[9pt] &= \frac{x^{12}}{12} + \frac{x^{13}}{13}. \end{align*}

Then, applying the recurrence,

    \begin{align*} &&11 \cdot F(10,2) + 2 \cdot F(11,1) &= x^{11} (1+x)^2 \\[9pt] \implies && 11 \cdot F(10,2) + 2 \left( \frac{x^{12}}{12} + \frac{x^{13}}{13} \right) &= x^{13} + 2x^{12} + x^{11} \\[9pt] \implies && 11 \cdot F(10,2) &= \frac{11\cdot x^{13}}{13} + \frac{11 \cdot x^{12}}{6} + x^{11} \\[9pt] \implies && F(10,2) &= \frac{x^{13}}{13} + \frac{x^{12}}{6} + \frac{x^{11}}{11}. \end{align*}

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):