Definite IntegrationmediumFree

Definite Integration: Prime Equal Denotes Fractional Part

JEE Maths question with a full step-by-step solution.

Question

\frac{1}{e-1}\sum_{n=1}^{50}\frac{1}{(n+1)!}\int_{-n}^{n}e^{\{|x|\}}\,\frac{dx}{1+e^{\sin x}} = 1-\frac{1}{(ab)!}

where

a

and

b

are prime, then

\left|a-b\right|

is equal to (

\{\,\cdot\,\}

denotes the fractional part)

Solution

Answer: 14 (± 0.01)

Step 1: For any even function

g

and any odd function

h

\int_{-n}^{n}\frac{g(x)}{1+e^{h(x)}}\,dx = \int_{0}^{n}g(x)\,dx

Proof: add the integral to its image under

x \to -x

; the two denominators

1+e^{h}

and

1+e^{-h}

have reciprocals summing to

1

. Step 2:

\sin x

is odd, and

e^{\{|x|\}}

depends on

|x|

, so it is even. Hence

\int_{-n}^{n}e^{\{|x|\}}\frac{dx}{1+e^{\sin x}} = \int_{0}^{n}e^{\{x\}}\,dx

Step 3:

\{x\}

has period

1

, so

e^{\{x\}}

does too, and over

\left[0,n\right]

the integral is

n

copies of the integral over one period:

\int_{0}^{n}e^{\{x\}}\,dx = n\int_{0}^{1}e^{x}\,dx = n\left(e-1\right)

Step 4:

\frac{1}{e-1}\sum_{n=1}^{50}\frac{n\left(e-1\right)}{(n+1)!} = \sum_{n=1}^{50}\frac{n}{(n+1)!}

Step 5:

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

\sum_{n=1}^{50}\left[\frac{1}{n!}-\frac{1}{(n+1)!}\right] = \frac{1}{1!}-\frac{1}{51!} = 1-\frac{1}{51!}

Step 6:

a

and

b

are prime, so

ab \ge 4

, and

m!

is strictly increasing for

m \ge 2

. Hence

\left(ab\right)! = 51! \quad\Longrightarrow\quad ab = 51 = 3\times17

3\times17

is the prime factorisation of

51

, so the only pair of primes with product

51

\{a,b\} = \{3,17\}

, and

\left|a-b\right| = \left|3-17\right| = 14

Answer:

14.00

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