Integration

16 methods · 45 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 $\int_{0}^{1} x^{x} d x = 1 - (\frac{1}{2})^{2} + (\frac{1}{3})^{3} - (\frac{1}{4})^{4} + \dots .$

Worked solution

Part (i).

We substitute

x = e^{- t}

, so that as

x \to 0^{+}

we have

t \to + \infty

, and

ln x = - t

. Then

x^{m} (ln x)^{n} = e^{- m t} (- t)^{n} = (- 1)^{n} e^{- m t} t^{n} .

Since

t \to \infty lim e^{- m t} t^{n} = 0

(exponential decay beats any polynomial), it follows immediately that

x \to 0^{+} lim x^{m} (ln x)^{n} = 0.

For the second part, write

x^{x} = e^{x l n x}

. We need the exponent

x ln x

x \to 0^{+}

. Taking

m = n = 1

in the result above gives

x \to 0^{+} lim x ln x = 0

. Since the exponential function is continuous,

x \to 0^{+} lim x^{x} = x \to 0^{+} lim e^{x l n x} = e^{0} = 1.

Part (ii).

To find the reduction formula for

I_{n} = \int_{0}^{1} x^{m} (ln x)^{n} d x

, integrate by parts. Set

u = (ln x)^{n}

and

d v = x^{m} d x

, giving

d u = \frac{n ( ln x ) ^{n - 1}}{x} d x

and

v = \frac{x ^{m + 1}}{m + 1}

. Then

I_{n} = [\frac{x ^{m + 1}}{m + 1} (ln x)^{n}]_{0}^{1} - \int_{0}^{1} \frac{x ^{m + 1}}{m + 1} \cdot \frac{n ( ln x ) ^{n - 1}}{x} d x .

The boundary term vanishes: at

x = 1

ln 1 = 0

; at

x = 0^{+}

x^{m + 1} (ln x)^{n} \to 0

by part (i). Hence

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

Shifting the index (replacing

n

n + 1

) gives

I_{n + 1} = - \frac{n + 1}{m + 1} I_{n} .

Applying the recurrence repeatedly:

I_{n} = - \frac{n}{m + 1} I_{n - 1} = (- \frac{1}{m + 1})^{2} n (n - 1) I_{n - 2} = \dots = (- \frac{1}{m + 1})^{n} n! I_{0} .

Since

I_{0} = \int_{0}^{1} x^{m} d x = \frac{1}{m + 1}

, we obtain

I_{n} = \frac{( - 1 ) ^{n} n !}{( m + 1 ) ^{n + 1}} .

Part (iii).

Expand

x^{x} = e^{x l n x}

as a power series:

x^{x} = k = 0 \sum \infty \frac{( x ln x ) ^{k}}{k !} .

Integrating term by term over

[0, 1]

(valid since all terms are bounded and the series converges uniformly on

(0, 1]

\int_{0}^{1} x^{x} d x = k = 0 \sum \infty \frac{1}{k !} \int_{0}^{1} x^{k} (ln x)^{k} d x = k = 0 \sum \infty \frac{I _{k}^{(m = k)}}{k !},

where

I_{k}^{(m = k)}

uses the result from part (ii) with

m = k

\int_{0}^{1} x^{k} (ln x)^{k} d x = \frac{( - 1 ) ^{k} k !}{( k + 1 ) ^{k + 1}} .

Therefore

\int_{0}^{1} x^{x} d x = k = 0 \sum \infty \frac{( - 1 ) ^{k} k !}{k ! ( k + 1 ) ^{k + 1}} = k = 0 \sum \infty \frac{( - 1 ) ^{k}}{( k + 1 ) ^{k + 1}} .

Re-indexing with

j = k + 1

\int_{0}^{1} x^{x} d x = j = 1 \sum \infty \frac{( - 1 ) ^{j - 1}}{j ^{j}} = 1 - (\frac{1}{2})^{2} + (\frac{1}{3})^{3} - (\frac{1}{4})^{4} + \dots

as required.

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 $I_{\frac{3}{2}}$ .

Worked solution

Part (i).

For the first symmetry, use the substitution

x = 1 - t

, so

d x = - d t

, and when

x = 0

t = 1

; when

x = 1

t = 0

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

Adding the original integral to itself after this flip:

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

For the recurrence, integrate

I_{n}

by parts: write

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

and integrate

(1 - x)^{n}

to get

- \frac{( 1 - x ) ^{n + 1}}{n + 1}

I_{n} = [- \frac{x ^{n} ( 1 - x ) ^{n + 1}}{n + 1}]_{0}^{1} + \frac{n}{n + 1} \int_{0}^{1} x^{n - 1} (1 - x)^{n + 1} d x = \frac{n}{n + 1} \int_{0}^{1} x^{n - 1} (1 - x)^{n + 1} d x .

Now split

(1 - x)^{n + 1} = (1 - x)^{n} - x (1 - x)^{n}

\frac{n}{n + 1} \int_{0}^{1} x^{n - 1} (1 - x)^{n + 1} d x = \frac{n}{n + 1} [\int_{0}^{1} x^{n - 1} (1 - x)^{n} d x - \int_{0}^{1} x^{n} (1 - x)^{n} d x] = \frac{n}{n + 1} [\frac{I _{n - 1}}{2} - I_{n}] .

I_{n} = \frac{n}{n + 1} \cdot \frac{I _{n - 1}}{2} - \frac{n}{n + 1} I_{n}

, giving

I_{n} (1 + \frac{n}{n + 1}) = \frac{n}{2 ( n + 1 )} I_{n - 1}

, i.e.

\frac{2 n + 1}{n + 1} I_{n} = \frac{n}{2 ( n + 1 )} I_{n - 1}

, and therefore:

I_{n} = \frac{n}{2 ( 2 n + 1 )} I_{n - 1} .

Part (ii).

Proceed by induction. For

n = 1

I_{0} = \int_{0}^{1} 1 d x = 1

and

I_{1} = \frac{1}{2 \cdot 3} I_{0} = \frac{1}{6}

. Also

\frac{( 1 ! ) ^{2}}{3 !} = \frac{1}{6}

. Base case holds.

Assume

I_{n - 1} = \frac{(( n - 1 )! ) ^{2}}{( 2 n - 1 )!}

. Then by the recurrence:

I_{n} = \frac{n}{2 ( 2 n + 1 )} \cdot \frac{(( n - 1 )! ) ^{2}}{( 2 n - 1 )!} = \frac{n \cdot (( n - 1 )! ) ^{2}}{2 ( 2 n + 1 ) ( 2 n - 1 )!} .

Note

n \cdot ((n - 1)!)^{2} = n! (n - 1)!

and

2 (2 n + 1) (2 n - 1)! = \frac{( 2 n + 1 )!}{( 2 n - 1 )!! / ( 2 n - 1 )!!}

… more cleanly:

2 (2 n + 1) (2 n - 1)! = (2 n + 1)! / ((2 n - 1)!! \cdot 2^{n - 1} / (2^{n - 1}))

. Directly: multiply numerator and denominator by

2 n

I_{n} = \frac{n ! ( n - 1 )!}{2 ( 2 n + 1 ) ( 2 n - 1 )!} \cdot \frac{2 n}{2 n} = \frac{n ! \cdot ( n !) \cdot \frac{1}{n} \cdot 2 n}{2 n ( 2 n + 1 ) ( 2 n - 1 )!} .

Simpler:

n \cdot ((n - 1)!)^{2} = n! (n - 1)!

and

2 (2 n + 1) (2 n - 1)! = (2 n + 1)! / (2 n \cdot (2 n - 2) \dots 2)

. Observe

2 (2 n + 1) (2 n - 1)! = (2 n + 1)! / (2 n \cdot (2 n - 2) \dots 2)

is cumbersome; instead use

(2 n + 1)! = (2 n + 1) \cdot 2 n \cdot (2 n - 1)!

and insert

2 n

in both numerator and denominator:

I_{n} = \frac{n \cdot (( n - 1 )! ) ^{2}}{2 ( 2 n + 1 ) ( 2 n - 1 )!} = \frac{( n ! ) ^{2}}{2 n \cdot 2 ( 2 n + 1 ) ( 2 n - 1 )!} \cdot 2 n \cdot \frac{1}{n} \cdot n = \frac{( n ! ) ^{2}}{( 2 n + 1 )!} .

Indeed

2 n \cdot (2 n + 1) \cdot (2 n - 1)! = (2 n + 1)!

, so

I_{n} = \frac{n ! ( n - 1 )!}{2 ( 2 n + 1 ) ( 2 n - 1 )!} = \frac{n ! \cdot n !}{2 n \cdot 2 ( 2 n + 1 ) ( 2 n - 1 )!} \cdot 2 = \frac{( n ! ) ^{2}}{( 2 n + 1 )!} .

By induction,

I_{n} = \frac{( n ! ) ^{2}}{( 2 n + 1 )!}

for all positive integers

n

.

Part (iii).

Set

x = sin^{2} θ

, so

d x = 2 sin θ cos θ d θ

, and

1 - x = cos^{2} θ

. The limits

x = 0, 1

give

θ = 0, \frac{π}{2}

I_{1/2} = \int_{0}^{π /2} (sin^{2} θ)^{1/2} (cos^{2} θ)^{1/2} \cdot 2 sin θ cos θ d θ = \int_{0}^{π /2} 2 sin^{2} θ cos^{2} θ d θ = \frac{1}{2} \int_{0}^{π /2} sin^{2} (2 θ) d θ = \frac{1}{2} \cdot \frac{π}{4} = \frac{π}{8} .

For

I_{3/2}

, apply the recurrence from part (i) with

n = \frac{3}{2}

I_{3/2} = \frac{\frac{3}{2}}{2 ( 2 \cdot \frac{3}{2} + 1 )} I_{1/2} = \frac{\frac{3}{2}}{2 \cdot 4} \cdot \frac{π}{8} = \frac{3}{64} \cdot \frac{π}{1} = \frac{3 π}{64} \cdot \frac{1}{1} .

More carefully:

\frac{n}{2 ( 2 n + 1 )}

n = \frac{3}{2}

gives

\frac{3/2}{2 \cdot 4} = \frac{3}{16}

, so

I_{3/2} = \frac{3}{16} \cdot \frac{π}{8} = \frac{3 π}{128} .

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.

Worked solution

Part (i) — Recurrence on $I_{n}$ .

We write

\frac{1}{( 1 + u ^{2} ) ^{n}} = \frac{1}{( 1 + u ^{2} ) ^{n + 1}} + \frac{u ^{2}}{( 1 + u ^{2} ) ^{n + 1}}

, which immediately gives

I_{n} - I_{n + 1} = \int_{0}^{\infty} \frac{u ^{2}}{( 1 + u ^{2} ) ^{n + 1}} d u

.

We integrate this last integral by parts: we set

u^{'} = u

and

v = \frac{1}{( 1 + u ^{2} ) ^{n + 1}}

, so

v^{'} = - \frac{( 2 n + 2 ) u}{2 ( 1 + u ^{2} ) ^{n + 2}} = - \frac{( n + 1 ) u}{( 1 + u ^{2} ) ^{n + 2}}

(or alternatively we use

u = u \cdot 1

and integrate

\frac{u}{( 1 + u ^{2} ) ^{n + 1}}

). More directly: setting

u = tan θ

I_{n} - I_{n + 1} = \int_{0}^{\infty} \frac{u ^{2}}{( 1 + u ^{2} ) ^{n + 1}} d u .

Integration by parts with

p = u

d q = \frac{u}{( 1 + u ^{2} ) ^{n + 1}} d u

: we obtain an integrable

d q

(

q = - \frac{1}{2 n ( 1 + u ^{2} ) ^{n}}

), and the boundary term vanishes at both endpoints. This leaves

I_{n} - I_{n + 1} = \frac{1}{2 n} \int_{0}^{\infty} \frac{1}{( 1 + u ^{2} ) ^{n}} d u = \frac{1}{2 n} I_{n} .

Hence

I_{n + 1} = (1 - \frac{1}{2 n}) I_{n} = \frac{2 n - 1}{2 n} I_{n}

.

Since

I_{1} = \frac{π}{2}

(a classical integral), we iterate:

I_{n + 1} = \frac{2 n - 1}{2 n} \cdot \frac{2 n - 3}{2 n - 2} \dots \frac{1}{2} \cdot \frac{π}{2} .

Multiplying numerator and denominator by

2^{n} \cdot n!

to bring out factorials:

I_{n + 1} = \frac{( 2 n )!}{2 ^{2 n} ( n ! ) ^{2}} \cdot \frac{π}{2} = \frac{( 2 n )! π}{2 ^{2 n + 1} ( n ! ) ^{2}} .

Part (ii) — The substitution $x \mapsto x^{- 1}$ .

The key idea:

(x - x^{- 1})^{2}

is invariant under

x \mapsto x^{- 1}

. Performing the change

x = t^{- 1}

J = \int_{0}^{\infty} f ((x - x^{- 1})^{2}) d x

(with

d x = - t^{- 2} d t

, and the bounds swapped), we obtain

J = \int_{0}^{\infty} x^{- 2} f ((x - x^{- 1})^{2}) d x .

Adding the two expressions for

J

gives

2 J = \int_{0}^{\infty} (1 + x^{- 2}) f ((x - x^{- 1})^{2}) d x

, so

J = \frac{1}{2} \int_{0}^{\infty} (1 + x^{- 2}) f ((x - x^{- 1})^{2}) d x

.

Finally, we set

u = x - x^{- 1}

. Then

d u = (1 + x^{- 2}) d x

. As

x

ranges over

(0, + \infty)

u

ranges over

(- \infty, + \infty)

, so by symmetry of

f (u^{2})

J = \frac{1}{2} \int_{- \infty}^{+ \infty} f (u^{2}) d u = \int_{0}^{\infty} f (u^{2}) d u .

Part (iii) — Evaluating the integral.

We identify

f

by rewriting the integrand. Note that

x^{4} - x^{2} + 1 = x^{2} (x^{2} - 1 + x^{- 2}) = x^{2} ((x - x^{- 1})^{2} + 1)

, so

\frac{x ^{2 n - 2}}{( x ^{4} - x ^{2} + 1 ) ^{n}} = \frac{x ^{2 n - 2}}{x ^{2 n} ( ( x - x ^{- 1} ) ^{2} + 1 ) ^{n}} = \frac{x ^{- 2}}{( ( x - x ^{- 1} ) ^{2} + 1 ) ^{n}} .

This matches

x^{- 2} f ((x - x^{- 1})^{2})

with

f (t) = \frac{1}{( 1 + t ) ^{n}}

. By part (ii), this integral equals

J = \int_{0}^{\infty} f (u^{2}) d u = \int_{0}^{\infty} \frac{1}{( 1 + u ^{2} ) ^{n}} d u = I_{n} = I_{(n - 1) + 1}

.

Applying the formula from part (i) with

n

replaced by

n - 1

\int_{0}^{\infty} \frac{x ^{2 n - 2}}{( x ^{4} - x ^{2} + 1 ) ^{n}} d x = I_{n} = \frac{( 2 n - 2 )! π}{2 ^{2 n - 1} ( ( n - 1 )! ) ^{2}} .