Integration

11 methods · 43 questions

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When an integral depends on an integer (or half-integer) parameter $n$ , do not attack it directly: relate $I_{n}$ to a lower case. The STEP-specific moves are (a) form $I_{n} - I_{n - 1}$ (or $I_{n + 1} + c I_{n}$ ) so that the difference collapses to something elementary which then telescopes, and (b) integrate by parts after splitting one power off, often using an identity ( $sin^{2} = 1 - cos^{2}$ , $sech^{n + 2} sinh^{2}$ ) to relate $I_{n}$ to $I_{n \pm 2}$ .

Trigger: a family $I_{n}$ , especially with a 'show that $I_{n} - I_{n - 1} = \dots$ ' or 'find a reduction formula' instruction. Once you have the recurrence, either iterate to a closed form or sum a telescoping series. The reusable insight is that the difference of consecutive members is the engine, even when the question only names the original $I_{n}$ .

Linked questions (3)

STEP 3 2009 · Q8

Let

m

be a positive integer and let

n

be a non-negative integer.

Use the result $t \to \infty lim e^{- m t} t^{n} = 0$ to show that $x \to 0 lim x^{m} (ln x)^{n} = 0 .$ By writing $x^{x}$ as $e^{x l n x}$ show that $x \to 0 lim x^{x} = 1 .$
Let $I_{n} = \int_{0}^{1} x^{m} (ln x)^{n} d x$ . Show that $I_{n + 1} = - \frac{n + 1}{m + 1} I_{n}$ and hence evaluate $I_{n}$ .
Show that $\int_{0}^{1} x^{x} d x = 1 - (\frac{1}{2})^{2} + (\frac{1}{3})^{3} - (\frac{1}{4})^{4} + \dots .$

STEP 2 2013 · Q2

For

n ⩾ 0

, let

I_{n} = \int_{0}^{1} x^{n} (1 - x)^{n} d x .

For $n ⩾ 1$ , show by means of a substitution that $\int_{0}^{1} x^{n - 1} (1 - x)^{n} d x = \int_{0}^{1} x^{n} (1 - x)^{n - 1} d x$ and deduce that $2 \int_{0}^{1} x^{n - 1} (1 - x)^{n} d x = I_{n - 1} .$ Show also, for $n ⩾ 1$ , that $I_{n} = \frac{n}{n + 1} \int_{0}^{1} x^{n - 1} (1 - x)^{n + 1} d x$ and hence that $I_{n} = \frac{n}{2 ( 2 n + 1 )} I_{n - 1} .$
When $n$ is a positive integer, show that $I_{n} = \frac{( n ! ) ^{2}}{( 2 n + 1 )!} .$
Use the substitution $x = sin^{2} θ$ to show that $I_{\frac{1}{2}} = \frac{π}{8}$ , and evaluate $I_{\frac{3}{2}}$ .

STEP 3 2015 · Q1

Let $I_{n} = \int_{0}^{\infty} \frac{1}{( 1 + u ^{2} ) ^{n}} d u,$ where $n$ is a positive integer. Show that $I_{n} - I_{n + 1} = \frac{1}{2 n} I_{n}$ and deduce that $I_{n + 1} = \frac{( 2 n )! π}{2 ^{2 n + 1} ( n ! ) ^{2}} .$
Let $J = \int_{0}^{\infty} f ((x - x^{- 1})^{2}) d x,$ where $f$ is any function for which the integral exists. Show that $J = \int_{0}^{\infty} x^{- 2} f ((x - x^{- 1})^{2}) d x = \frac{1}{2} \int_{0}^{\infty} (1 + x^{- 2}) f ((x - x^{- 1})^{2}) d x = \int_{0}^{\infty} f (u^{2}) d u .$
Hence evaluate $\int_{0}^{\infty} \frac{x ^{2 n - 2}}{( x ^{4} - x ^{2} + 1 ) ^{n}} d x,$ where $n$ is a positive integer.