The Cardinal Sine Evaluation - Part 2

Calculus Level 4

$\large \displaystyle \int _{ 0 }^{ \frac { \pi }{ 2 } }{ f\left( x \right) f\left( \frac { \pi }{ 2 } -x \right) } \, dx$

In mathematics the cardinal sine function or $\text{sinc}(x)$ is defined as $f(x) = \frac {\sin(x)}x$ .

Which of these answer choices represent the value of above expression?

This problem is orignal. Try the sister problem: The Cardinal Sine Evaluation .

None of these choices

\frac { \pi }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ f\left( x \right) \, dx }

\frac { \pi }{ 4 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ f\left( x \right) \, dx }

\pi \int _{ 0 }^{ \pi }{ f\left( x \right) \, dx }

\frac { 2 }{ \pi } \int _{ 0 }^{ \pi }{ f\left( x \right) \, dx }

\frac { 2 }{ \pi } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ f\left( x \right)\, dx }

\frac { \pi }{ 2 } \int _{ 0 }^{ \pi }{ f\left( x \right) \, dx }

\pi \int _{ 0 }^{ \frac { \pi }{ 2 } }{ f\left( x \right) \, dx }

1 solution

Anupam Khandelwal
Jun 4, 2015

$The\quad integral\quad can\quad be\quad written\quad as\\ I\quad =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sin { (x) } \cos { (x) } }{ x(\frac { \pi }{ 2 } -x) } dx } \\ I\quad =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sin { (2x) } }{ x(\pi -2x) } .dx\\ } \\ Putting\quad 2x\quad =\quad t\quad we\quad get,\\ \\ I\quad =\int _{ 0 }^{ \pi }{ \frac { \sin { (t) } }{ t(\pi -t) } .dt } \quad \\ I\quad =\frac { 1 }{ \pi } \int _{ 0 }^{ \pi }{ \left( \frac { sin(t) }{ t } +\frac { sin(t) }{ \pi -t } \right) .dt } \\ Using\quad \int _{ 0 }^{ a }{ f\left( x \right) .dx } =\int _{ 0 }^{ a }{ f(a-x).dx } \\ The\quad above\quad expression\quad can\quad be\quad written\quad as\\ \quad \quad \quad \quad \quad \quad \quad \boxed { I\quad =\frac { 2 }{ \pi } \int _{ 0 }^{ \pi }{ f\left( x \right) .dx } } \\$

The Cardinal Sine Evaluation - Part 2

This problem is orignal. Try the sister problem: The Cardinal Sine Evaluation .

1 solution

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