Show that ${I_0} = \frac{1}{2}(1 + {{\text{e}}^{ - \pi }})$ .

By letting $y = x - n\pi$ , show that ${I_n} = {{\text{e}}^{ - n\pi }}{I_0}$ .

Hence determine the exact value of $\int_0^\infty  {{{\text{e}}^{ - x}}|\sin x|{\text{d}}x}$ .

${I_0} = \int_0^\pi  {{{\text{e}}^{ - x}}\sin x{\text{d}}x}$     M1

Note: Award M1 for ${I_0} = \int_0^\pi  {{{\text{e}}^{ - x}}|\sin x|{\text{d}}x}$

Attempt at integration by parts, even if inappropriate modulus signs are present.     M1

$= - \left[ {{{\text{e}}^{ - x}}\cos x} \right]_0^\pi - \int_0^\pi {{{\text{e}}^{ - x}}\cos x{\text{d}}x}$ or $= - \left[ {{{\text{e}}^{ - x}}\sin x} \right]_0^\pi - \int_0^\pi {{{\text{e}}^{ - x}}\cos x{\text{d}}x}$     A1

$= - \left[ {{{\text{e}}^{ - x}}\cos x} \right]_0^\pi - \left[ {{{\text{e}}^{ - x}}\sin x} \right]_0^\pi - \int_0^\pi {{{\text{e}}^{ - x}}\sin x{\text{d}}x}$ or $= - \left[ {{{\text{e}}^{ - x}}\sin x + {{\text{e}}^{ - x}}\cos x} \right]_0^\pi - \int_0^\pi {{{\text{e}}^{ - x}}\sin x{\text{d}}x}$     A1

$= - \left[ {{{\text{e}}^{ - x}}\cos x} \right]_0^\pi - \left[ {{{\text{e}}^{ - x}}\sin x} \right]_0^\pi - {I_0}$ or $- \left[ {{{\text{e}}^{ - x}}\sin x + {{\text{e}}^{ - x}}\cos x} \right]_0^\pi - {I_0}$     M1

Note: Do not penalise absence of limits at this stage

${I_0} = {{\text{e}}^{ - \pi }} + 1 - {I_0}$     A1

${I_0} = \frac{1}{2}(1 + {{\text{e}}^{ - \pi }})$     AG

Note: If modulus signs are used around cos x , award no accuracy marks but do not penalise modulus signs around sin x .

[6 marks]

${I_n} = \int_{n\pi }^{(n + 1)\pi } {{{\text{e}}^{ - x}}|\sin x|{\text{d}}x}$

Attempt to use the substitution $y = x - n\pi$     M1

(putting $y = x - n\pi$ , ${\text{d}}y = {\text{d}}x$ and $\left[ {n\pi ,{\text{ }}(n + 1)\pi } \right] \to [0,{\text{ }}\pi ]$)

so ${I_n} = \int_0^\pi  {{{\text{e}}^{ - (y + n\pi )}}|\sin (y + n\pi )|{\text{d}}y}$     A1

$= {{\text{e}}^{ - n\pi }}\int_0^\pi  {{{\text{e}}^{ - y}}|\sin (y + n\pi )|{\text{d}}y}$     A1

$= {{\text{e}}^{ - n\pi }}\int_0^\pi  {{{\text{e}}^{ - y}}\sin y{\text{d}}y}$     A1

$= {{\text{e}}^{ - n\pi }}{I_0}$     AG

[4 marks]

$\int_0^\infty  {{{\text{e}}^{ - x}}|\sin x|{\text{d}}x} = \sum\limits_{n = 0}^\infty  {{I_n}}$     M1

$= \sum\limits_{n = 0}^\infty  {{{\text{e}}^{ - n\pi }}{I_0}}$     (A1)

the $\sum$ term is an infinite geometric series with common ratio ${{\text{e}}^{ - \pi }}$     (M1)

therefore

$\int_0^\infty  {{{\text{e}}^{ - x}}|\sin x|{\text{d}}x} = \frac{{{I_0}}}{{1 - {{\text{e}}^{ - \pi }}}}$     (A1)

$= \frac{{1 + {{\text{e}}^{ - \pi }}}}{{2(1 - {{\text{e}}^{ - \pi }})}}{\text{ }}\left( { = \frac{{{{\text{e}}^\pi } + 1}}{{2({{\text{e}}^\pi } - 1)}}} \right)$     A1

[5 marks]

Part (a) is essentially core work requiring repeated integration by parts and many candidates realised that. However, some candidates left the modulus signs in ${I_0}$ which invalidated their work. In parts (b) and (c) it was clear that very few candidates had a complete understanding of the significance of the modulus sign and what conditions were necessary for it to be dropped. Overall, attempts at (b) and (c) were disappointing with few correct solutions seen.

Part (a) is essentially core work requiring repeated integration by parts and many candidates realised that. However, some candidates left the modulus signs in ${I_0}$ which invalidated their work. In parts (b) and (c) it was clear that very few candidates had a complete understanding of the significance of the modulus sign and what conditions were necessary for it to be dropped. Overall, attempts at (b) and (c) were disappointing with few correct solutions seen.

Part (a) is essentially core work requiring repeated integration by parts and many candidates realised that. However, some candidates left the modulus signs in ${I_0}$ which invalidated their work. In parts (b) and (c) it was clear that very few candidates had a complete understanding of the significance of the modulus sign and what conditions were necessary for it to be dropped. Overall, attempts at (b) and (c) were disappointing with few correct solutions seen.