One way of doing it is going the pure pythonical way, just define `f`

to be a python native function. For instance:

```
sage: var( 'x' );
sage: g(x) = x*sin(x)
sage: def f( n, g ): return g(x)^n
sage: f(3,g).integrate( x, 0, pi )
-40/9*pi + 2/3*pi^3
```

(The above code is mixing things, bad style, but `def f`

answers the question. There are also other ways, but we need the special situation...)

LATER EDIT since *There isn't any sort of conditional statement in the above code*
(Thanks for the remark.)

Above there is the $n$ as a power decorating $g$, not as a denominator, so that the contribution of $n$ cannot be simply moved mathematically using $\int_0^1 g(x)/k(n)\; dx =(1/k(n))\int_0^1 g(x)\, dx$.

But ok, let us require more examples using

```
def k(n): return (2 if n == 1 else n)
g(x) = x^3
```

We have for instance as in the example above, replacing the `3`

with `k(n)`

, thus using `f( k(n), g )`

. This was the answer to the comment, and we may stop here. But let us be more explicit in some examples with
$$ n\in{\ -1,\ 1, 2,\ 17\ }\ . $$

```
sage: NVALUES = [ -1,1,2,17 ]
sage: [ k(n) for n in NVALUES ]
[-1, 2, 2, 17]
sage: [ g(x)/k(n) for n in NVALUES ]
[-x^3, 1/2*x^3, 1/2*x^3, 1/17*x^3]
sage: def f(k,g): return g(x)/k
sage: [ f( k(n), g ) for n in NVALUES ]
[-x^3, 1/2*x^3, 1/2*x^3, 1/17*x^3]
sage: [ f( k(n), g ).integrate(x,0,1) for n in NVALUES ]
[-1/4, 1/8, 1/8, 1/68]
```

(One can of course move the `k(n)`

into the definition of `f`

.)

Note: The above solution is (rather less arguably) simpler than

```
sage: var('n');
sage: k = piecewise( [ ([1,1], 2), ((-Infinity,1),n), ((1,Infinity), n) ], var=n );
sage: k(n)
piecewise(n|-->2 on {1}, n|-->n on (-oo, 1), n|-->n on (1, +oo); n)
sage: f(x) = x^3 / k(n)
sage: f(n=-1).integrate( x,0,1 )
-1/4
sage: f(n=1).integrate( x,0,1 )
1/8
sage: f(n=2).integrate( x,0,1 )
1/8
sage: f(n=17).integrate( x,0,1 )
1/68
```

where we also insist to use symbolic expressions and the substitution `f(n=1)`

(into `f`

).