Evaluate this index Integral containing Trig

[bugatti79]

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?

 

[SammyS]

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

 

[bugatti79]

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...

 

[vela]

You can't integrate it like that when i=j. Assume i=j, simplify the integrand, and then integrate.

 

[SammyS]

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

 

[bugatti79]

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