How to get a universal answer using Integrate [duplicate]

Question

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 1

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}]]]