• Support PF! Buy your school textbooks, materials and every day products Here!

Evaluate this index Integral containing Trig

  • Thread starter bugatti79
  • Start date
  • #1
719
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 dont wrong?
 

Answers and Replies

  • #2
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,224
947

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
719
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 dont 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 dont know how that answer ##\frac{3 \pi^2}{4}## was evaluated...
 
  • #4
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,538
1,149
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
11,224
947
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
719
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
 

Related Threads for: Evaluate this index Integral containing Trig

  • Last Post
Replies
6
Views
826
  • Last Post
Replies
7
Views
1K
Replies
2
Views
438
  • Last Post
Replies
5
Views
663
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
16
Views
2K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
2
Views
786
Top