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Question: If \(\phi \left( t \right) = \left\\{ {1,{\text{ }}0 \leqslant t \leqslant 1} \right.\) ...

If \phi \left( t \right) = \left\\{ {1,{\text{ }}0 \leqslant t \leqslant 1} \right.
{\text{ = }}\left\\{ {{\text{0, otherwise}}} \right.,
Then ∫−30003000(∑r′=20142016ϕ(t−r′)ϕ(t−2016))dt=\int_{ - 3000}^{3000} {\left( {\sum\limits_{r' = 2014}^{2016} {\phi \left( {t - r'} \right)\phi \left( {t - 2016} \right)} } \right)dt} =
(a)\left( a \right) A real number
(b)1\left( b \right)1
(c)0\left( c \right)0
(d)\left( d \right) Does not exist

Explanation

Solution

In this particular question use the concept of splitting the definite integral into two or more than two parts according to their integration limits which is given as, ∫x=ax=bf(x)dx=∫x=ax=cf(x)dx+∫x=cx=bf(x)dx\int_{x = a}^{x = b} {f\left( x \right)dx} = \int_{x = a}^{x = c} {f\left( x \right)dx} + \int_{x = c}^{x = b} {f\left( x \right)dx} , where a<c<ba < c < b, so use these concepts to reach the solution of the question.

Complete step by step answer:
Given data:
If \phi \left( t \right) = \left\\{ {1,{\text{ }}0 \leqslant t \leqslant 1} \right.
{\text{ = }}\left\\{ {{\text{0, otherwise}}} \right.,
Then we have to find out the value of ∫−30003000(∑r′=20142016ϕ(t−r′)ϕ(t−2016))dt\int_{ - 3000}^{3000} {\left( {\sum\limits_{r' = 2014}^{2016} {\phi \left( {t - r'} \right)\phi \left( {t - 2016} \right)} } \right)dt}
Expand the summation we have,
⇒∫−30003000(ϕ(t−2014)ϕ(t−2016)+ϕ(t−2015)ϕ(t−2016)+ϕ(t−2016)ϕ(t−2016))dt\Rightarrow \int_{ - 3000}^{3000} {\left( {\phi \left( {t - 2014} \right)\phi \left( {t - 2016} \right) + \phi \left( {t - 2015} \right)\phi \left( {t - 2016} \right) + \phi \left( {t - 2016} \right)\phi \left( {t - 2016} \right)} \right)dt}
Now as we know that ∫x=ax=bf(x)dx=∫x=ax=cf(x)dx+∫x=cx=bf(x)dx\int_{x = a}^{x = b} {f\left( x \right)dx} = \int_{x = a}^{x = c} {f\left( x \right)dx} + \int_{x = c}^{x = b} {f\left( x \right)dx} , where a<c<ba < c < b, so use this property in the above equation we have,
⇒∫−30002016f(t)dt+∫20162017f(t)dt+∫20173000f(t)dt\Rightarrow \int_{ - 3000}^{2016} {f\left( t \right)dt} + \int_{2016}^{2017} {f\left( t \right)dt} + \int_{2017}^{3000} {f\left( t \right)dt}
Where, f(t)=ϕ(t−2014)ϕ(t−2016)+ϕ(t−2015)ϕ(t−2016)+ϕ(t−2016)ϕ(t−2016)f\left( t \right) = \phi \left( {t - 2014} \right)\phi \left( {t - 2016} \right) + \phi \left( {t - 2015} \right)\phi \left( {t - 2016} \right) + \phi \left( {t - 2016} \right)\phi \left( {t - 2016} \right)
Now as, \phi \left( t \right) = \left\\{ {1,{\text{ }}0 \leqslant t \leqslant 1} \right.
{\text{ = }}\left\\{ {{\text{0, otherwise}}} \right.,
So, f (t) = 0 in between (-3000, 2016) and in between (2017, 3000), so we have,
⇒∫−30003000(∑r′=20142016ϕ(t−r′)ϕ(t−2016))dt=∫−300020160dt+∫20162017f(t)dt+∫201730000dt\Rightarrow \int_{ - 3000}^{3000} {\left( {\sum\limits_{r' = 2014}^{2016} {\phi \left( {t - r'} \right)\phi \left( {t - 2016} \right)} } \right)dt} = \int_{ - 3000}^{2016} {0dt} + \int_{2016}^{2017} {f\left( t \right)dt} + \int_{2017}^{3000} {0dt}
Now as we know integration of zero is zero, so we have,
⇒∫−30003000(∑r′=20142016ϕ(t−r′)ϕ(t−2016))dt=∫20162017f(t)dt\Rightarrow \int_{ - 3000}^{3000} {\left( {\sum\limits_{r' = 2014}^{2016} {\phi \left( {t - r'} \right)\phi \left( {t - 2016} \right)} } \right)dt} = \int_{2016}^{2017} {f\left( t \right)dt}
Now f (t) = 1 in between (2016, 2017) so we have,
⇒∫−30003000(∑r′=20142016ϕ(t−r′)ϕ(t−2016))dt=∫201620171.dt\Rightarrow \int_{ - 3000}^{3000} {\left( {\sum\limits_{r' = 2014}^{2016} {\phi \left( {t - r'} \right)\phi \left( {t - 2016} \right)} } \right)dt} = \int_{2016}^{2017} {1.dt}
Now integrate it we have,
⇒∫−30003000(∑r′=20142016ϕ(t−r′)ϕ(t−2016))dt=[t]20162017\Rightarrow \int_{ - 3000}^{3000} {\left( {\sum\limits_{r' = 2014}^{2016} {\phi \left( {t - r'} \right)\phi \left( {t - 2016} \right)} } \right)dt} = \left[ t \right]_{2016}^{2017}
Now apply integration limits we have,
⇒∫−30003000(∑r′=20142016ϕ(t−r′)ϕ(t−2016))dt=[2017−2016]=1\Rightarrow \int_{ - 3000}^{3000} {\left( {\sum\limits_{r' = 2014}^{2016} {\phi \left( {t - r'} \right)\phi \left( {t - 2016} \right)} } \right)dt} = \left[ {2017 - 2016} \right] = 1
So this is the required answer.
Now as we all know that 1 is considered as a real number.

Hence options (a) and (b) are the correct answers.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall basic integration i.e. integration of zero is zero and integration of constant is the same variable by which we integrate, so apply this as above then apply integration limits we will get the required answer, and always recall that all integers from [−∞.+∞]\left[ { - \infty . + \infty } \right] are real numbers.