Evaluate this index Integral containing Trig

  • Thread starter bugatti79
  • Start date
  • #1
bugatti79
792
1

Homework Statement



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##

Homework Equations


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 don't wrong?
 

Answers and Replies

  • #2
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,698
1,275

Homework Statement



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##

Homework Equations


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 substitute ##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 done wrong?
Did you mean to say "numerator" ?

cos(0) = 1
 
  • #3
bugatti79
792
1

Homework Statement



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##

Homework Equations


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 don't wrong?

Did you mean to say "numerator" ?

cos(0) = 1

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 don't know how that answer ##\frac{3 \pi^2}{4}## was evaluated...
 
  • #4
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,764
2,401
You can't integrate it like that when i=j. Assume i=j, simplify the integrand, and then integrate.
 
  • #5
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,698
1,275
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 don't know how that answer ##\frac{3 \pi^2}{4}## was evaluated...
Well, if i = j then you are integrating cos((0)πx), but cos((0)πx) = 1
 
  • #6
bugatti79
792
1
You can't integrate it like that when i=j. Assume i=j, simplify the integrand, and then integrate.

Well, if i = j then you are integrating cos((0)πx), but cos((0)πx) = 1

Thank you, I thought one has to insert the index values AFTER integration...? Where is the mathematical rule that asserts this?

Cheers
 

Suggested for: Evaluate this index Integral containing Trig

  • Last Post
Replies
12
Views
349
  • Last Post
Replies
6
Views
481
Replies
10
Views
824
Replies
1
Views
263
  • Last Post
Replies
3
Views
431
Replies
3
Views
244
Replies
14
Views
726
Replies
5
Views
607
Top