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Evaluate this index Integral containing Trig

  1. Aug 29, 2012 #1
    1. The problem statement, all variables and given/known data

    Folks Evaluate ##B_{11}## given

    ##\displaystyle B_{ij}=\int_0^1 (1+x) \frac{d \phi_i}{dx} \frac{ d\phi_j}{dx} dx## where ##\phi_i= sin i \pi x## and ##\phi_j=sin j \pi x##

    2. Relevant equations
    3. The attempt at a solution

    I calculate ##\displaystyle B_{ij}=\int_0^1 (1+x)[ i \pi \cos(i \pi x))(j \pi \cos(j \pi x)]=ij \pi^2 \int_0^1 (1+x) \left[\frac{1}{2} [\cos(i+j)\pi x+\cos(i-j)\pi x\right ]dx##



    ##\displaystyle = \frac{ij \pi^{2}}{2} \int_0^1\left [ \cos(i+j) \pi x+\cos(i-j) \pi x +x \cos(i+j) \pi x+x \cos(i-j) \pi x \right ]dx##

    Now for the second and last term in the integrand if we substitue ##i=j=1## after integrating we will get a 0 in the denominator ....but the book calculates ##B_{ij}=\frac{3 \pi^2}{4}##

    What have I dont wrong?
     
  2. jcsd
  3. Aug 29, 2012 #2

    SammyS

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    Did you mean to say "numerator" ?

    cos(0) = 1
     
  4. Aug 29, 2012 #3
    Well no...looking at the second term ##\displaystyle \int_0^1 \cos(i-j) \pi x dx= \left [\frac{sin(i-j) \pi x}{\pi(i-j)} \right]_0^1## 1 is indeterminate....

    I dont know how that answer ##\frac{3 \pi^2}{4}## was evaluated...
     
  5. Aug 29, 2012 #4

    vela

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    You can't integrate it like that when i=j. Assume i=j, simplify the integrand, and then integrate.
     
  6. Aug 29, 2012 #5

    SammyS

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    Well, if i = j then you are integrating cos((0)πx), but cos((0)πx) = 1
     
  7. Aug 29, 2012 #6
    Thank you, I thought one has to insert the index values AFTER integration....? Where is the mathematical rule that asserts this?

    Cheers
     
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