calculus and analysis - How to get a universal answer using Integrate

The following two codes give conflicting answers, when integrating $\sin(k\pi x)\sin(2n\pi x)$ from $0$ to $1$, where both $k$ and $n$ are positive integers.

Code 1 assumes that $k$,$n$ are independent integers:

Integrate[
  Sin[k*Pi*x]*Sin[2*n*Pi*x], {x, 0, 1}, 
  Assumptions -> {k ∈ Integers, n ∈ Integers}]

The result given by mathematica is 0.

Code 2 assumes that $k=2n$ and $n$ is integer:

Integrate[
  Sin[k*Pi*x]*Sin[2*n*Pi*x], {x, 0, 1},
  Assumptions -> {n ∈ Integers，k = 2*n}]

The result is 1/2.

The result of Code 2 should be included in that of Code 1. It seems that Code 1 doesn't manage to give a universal result. Isn't Code 1 supposed to give a universal result? if not, how to get one?

Answer

Mathematica works on the general case, not the specific, when it comes to simplifications. Let try to find what is going on. Mathematica says that

$$\begin{align*} I & =\int_{0}^{1}\sin\left( k\pi x\right) \sin\left( 2n\pi x\right) dx\\ & =\frac{1}{2\pi}\left( \frac{\sin\left( \left( k-2n\right) \pi\right) }{k-2n}-\frac{\sin\left( \left( k+2n\right) \pi\right) }{k+2n}\right) \end{align*}$$

Now, when you said that $k,n$ are integers, then $k-2n$ is also an integer, as well as $k+2n$. Therefore the above becomes zero. Which agrees with what Mathematica gives. Mathematica will not consider the special case here of what happenes if $k=2n$, since this is a special case of $k$.

Now lets look at what happens when you give specific case when $k=2n$. Then the result above becomes

$$\begin{align*} I & =\frac{1}{2\pi}\left( \frac{\sin\left( \left( 2n-2n\right) \pi\right) }{2n-2n}-\frac{\sin\left( \left( 2n+2n\right) \pi\right) }{2n+2n}\right) \\ & =\frac{1}{2\pi}\left( \frac{\sin\left( m\pi\right) }{m}-\frac{\sin\left( 4n\pi\right) }{4n}\right) \end{align*}$$

Where $m=2n-2n$. (did not want to put zero, since need to take limit). Then the above becomes

$$I=\frac{1}{2\pi}\frac{\sin\left( m\pi\right) }{m}-\frac{1}{2\pi}\frac {\sin\left( 4n\pi\right) }{4n}$$

Since $n$ is integer, then the second term above is zero. (notice also, here there is special case, what if $n=0$? Then you'll get 1/2 also for the second term and the whole thing becomes zero, like case 1, But since $n=0$ is special case, it is not considered). Now the above becomes

$$I=\frac{1}{2\pi}\frac{\sin\left( m\pi\right) }{m}$$

But $\lim_{m\rightarrow0}\frac{\sin m\pi}{m}=\pi$, hence the above becomes $$\begin{align*} I & =\frac{1}{2\pi}\pi\\ & =\frac{1}{2} \end{align*}$$

Which is what Mathematica gives.

ClearAll[n,k,x]
Assuming[ Element[k,Integers]&&Element[n,Integers],
          Simplify[Integrate[Sin[k Pi x]*Sin[2*n Pi x],{x,0,1}]]]
(*0*)



Assuming[ Element[n, Integers] && k == 2 n, 
     Simplify[Integrate[Sin[k Pi x]*Sin[2*n Pi x], {x, 0, 1}]]]