Question

Can someone check my answer - Integral Calculus Question

Answer 1

\[\large{\color{red}{\rm{Question:}}}\]

Let a be a fixed real number satisfying \(\large{0 < a < \pi}\) , and set:

\[\large{I_r = \int_{-a}^{a}\cfrac{1-r\cos u}{1-2r\cos u + r^2} \ du}\]

Prove that:

\[\large{I_1,~ \lim_{r \rightarrow 1^+} I_r ,~ \lim_{r \rightarrow 1^-} I_r}\]

all exist and are distinct.

Answer 2

\[\large{\color{red}{\rm{Hint:}}}\]

For r > 0 and \(\large{r \ne 1}\) show that:

\[\large{I_r = a + \cfrac{1}{2}(1-r^2)\int_{-a}^{a} \cfrac{du}{1 - 2r\cos u + r^2}}\]

and evaluate the integral using the transformation \(\large{t = \tan \cfrac{u}{2}}\).

Answer 3

\[\large{\color{red}{\rm{Solution: }}}\]


First I will calculate \(\large{I_1}\). We have:

\[\large{I_1 = \int_{-a}^{a} \cfrac{1-\cos u}{1-2\cos u + 1} \ du}\]
\[\large{\implies I_1 = \int_{-a}^{a} \cfrac{1-\cos u}{2-2\cos u} \ du}\]
\[\large{\implies I_1 = \int_{-a}^{a} \cfrac{1}{2} \ du = a}\tag{1}\]

where I have taken the value of the integrand to be 1/2 when u = 0.

Now, for r > 0 and \(\large{r \ne 1}\), we have:

\[\large{1-2r\cos u + r^2 > 2r - 2r\cos u = 2r(1-\cos u) \ge 0}\]

so that the integrand of the integral in equation (1) is continuous on [-a,a].

Now, we have:

\[\large{I_r = \int_{-a}^{a}\left[\cfrac{1}{2}+\cfrac{(1-r^2)}{2(1-2r\cos u+r^2)} \right] \ du}\]

\[\large{= a + \cfrac{1}{2}(1-r^2)\int_{-a}^{a} \cfrac{du}{1-2r\cos u + r^2}}\]

This gives:

\[\large{I_r = a+\cfrac{(1-r^2)}{2} \ J_r}\tag{2}\]

where:

\[\large{J_r = \int_{-a}^{a}\cfrac{du}{1-2r\cos u +r^2}}\tag{3}\]

Let \(\large{t = \tan {\cfrac{u}{2}}}\) with \(\large{-a \le u \le a}\) so that

\[\large{\cos u = \cfrac{1-t^2}{1+t^2}, \ du = \cfrac{2}{1+t^2} \ dt}\]

Using the above transformation in equation (3), we obtain:

\[\large{J_r = \cfrac{2}{(1+r)^2} \int_{-t_1}^{t_1} \cfrac{dt}{t^2 + \left(\cfrac{1-r}{1+r}\right)^2}}\tag{4}\]

where

\[\large{t_1 = \tan \cfrac{a}{2}}\]

I evaluated the standard integral in equation (4) and got:

\[\large{J_r = \cfrac{4}{\left|1-r^2 \right|}\tan^{-1} \left(\left|\cfrac{1+r}{1-r}\right| \tan \cfrac{a}{2}\right)}\tag{5}\]

From equations (2) and (5) we get:

\[\large{I_r = a + 2\cfrac{(1-r^2)}{|1-r^2|}\tan^{-1} \left(\left|\cfrac{1+r}{1-r}\right| \tan \cfrac{a}{2}\right)}\]

Hence for r > 1 we have:

\[\large{I_r = a - 2\tan^{-1} \left(\left|\cfrac{r+1}{r-1}\right|\tan \cfrac{a}{2}\right)}\]

and for 0 < r < 1 we have:

\[\large{I_r = a + 2\tan^{-1} \left(\left(\cfrac{1+r}{1-r}\right)\tan \cfrac{a}{2}\right)}\]

Taking limits we obtain:

\[\large{\lim_{r \rightarrow 1^+} I_r = a - 2 \cdot \cfrac{\pi}{2} = a - \pi}\]

\[\large{\lim_{r \rightarrow 1^-} I_r = a + 2\cdot \cfrac{\pi}{2} = a + \pi}\]

Thus the quantities \(\large{I_1,~\lim_{r \rightarrow 1^+} I_r,~\lim_{r \rightarrow 1^-} I_r}\) all exist and are all distinct.

Answer 4

@ganeshie8 @ikram002p @Kainui Can you please check this ? I have no correct answer available.

My solution is solely based on the `Hint` given in the book.

Answer 5

@ganeshie8 is it correct ?

Answer 6

the solution looks flawless to me xD

Answer 7

Thanks :)

I am surprised it worked :P ;)

Answer 8

its sounds like elliptic integral mmm

Answer 9

could u explan this statemennt :

```
where I have taken the value of the integrand to be 1/2 when u = 0.
```

the integral evaluates to 1/2 when r=1, right ?
whats with u=0 ?

Answer 10

Actually I had no other option :P

I was trying to define I_1 but couldn't define it for u = 0, so that is why I assumed it to be 1/2

I am stuck only in this step .... Any idea ?

Answer 11

oh :o
polars functions hey idk why im seeing cardiod

Answer 12

yeah looks this problem was cooked up from change of variables...

one sec vishy, im going thru it again :)

Answer 13

Sure take your time :)

Answer 14

i think we don't need to bother about the discontinuity...
\(\large I_1 = 1/2\) without any restrictions and moreover, u is just a dummy variable so it shouldn't matter

Answer 15

the solution looks real nice to me !
not sure if there is any shortcut...

Answer 16

Thanks a lot :)

Answer 17

I'm looking for shortcuts... So far all I have noticed is that the integral appears to be even so we can change the limits of integration to be from 0 to a and multiply the integral by 2.

The other thing I'm noticing is that the bottom appears to suspiciously look like the law of cosines.

Answer 18

Best of luck @Kainui :)

I am trying too but cannot figure anything :(