Need help with integrals from Landau-Lifshitz vol. 1 problems

[JevKus]

Hi!

I'm new to this forum and forums in general, so please be forgiving.

I am currently going through problems in Landau-Lifgarbagez's vol. 1 (Mechanics) and encountered two integrals I can't solve. The physical basis of the problem is crystal clear, but I can't do the final computation. The integrals are from problem 2 (b,c) par. 11. Mathematica spits out correct answers, so I won't post the full problem and will only post integrals I'm having problems with.

The first one is:
$\int_{1}^{a}\frac{dz}{z \sqrt{z^{2} - 1}\sqrt{a^{2} - z^{2}}}$;

here $z$ and $a>1$ are assumed to be real.

The other one is:
$\int_{0}^{a}\frac{dz}{(1 + z^{2})\sqrt{a^{2} - z^{2}}}$.

$z$ and $a$ are also assumed to be real.

Thank you!

 

[mighty2000]

Dear fellow forum member,

Welcome to the forum! I am happy to help with your integrals from Landau-Lifgarbagez's vol. 1. I understand that you are having trouble solving the integrals and that Mathematica is giving you the correct answers. However, as a scientist, it is important to understand the process and reasoning behind the solution rather than just relying on a computer program.

Let's take a look at the first integral:
\int_{1}^{a}\frac{dz}{z \sqrt{z^{2} - 1}\sqrt{a^{2} - z^{2}}}

One approach to solving this integral is by using the substitution u = z^2 - 1. This will transform the integral into:
\int_{0}^{a^2 - 1}\frac{du}{2u\sqrt{a^{2} - u}}

Now, we can use the substitution v = u/a^2. This will give us:
\int_{0}^{1}\frac{dv}{2v\sqrt{1 - v}}

This is a standard integral that can be solved using the trigonometric substitution v = sin^2(theta). After solving for v, we can substitute back in u and z to get our final solution.

Moving on to the second integral:
\int_{0}^{a}\frac{dz}{(1 + z^{2})\sqrt{a^{2} - z^{2}}}

We can use the same substitution u = z^2 - a^2, which will transform the integral into:
\int_{-a^2}^{0}\frac{-du}{2u\sqrt{-u}}

Using the substitution v = -u, we get:
\int_{0}^{a^2}\frac{dv}{2v\sqrt{v}}

Again, this is a standard integral that can be solved using the substitution v = sin^2(theta). After solving for v, we can substitute back in u and z to get our final solution.

I hope this helps in solving the integrals. Remember, as a scientist, it is important to understand the process and reasoning behind the solution rather than just relying on a computer program. Keep practicing and don't hesitate to ask for help if you encounter any more difficulties.