# Can integral be replaced by constant

1. Aug 28, 2014

### smslce

Let a,b are constants.
$$a=\int_{-∏}^∏ f(x)\,dx$$
and
$$b=\int_{-∏}^∏ {g(x)f(x)}\,dx,$$
Then by integration byparts, we have ,
b = $$(\left.g(x)\int_{-∏}^∏ f(x)\,dx)\right|_{-∏}^∏-\int_{-∏}^∏g'(x)\int_{-∏}^∏ f(x)\,dx \,dx$$

Is below step valid.
b = $$a(\left.g(x))\right|_{-∏}^∏-a\int_{-∏}^∏g'(x)\,dx$$

I am not able to find it googling. Can an integral be replaced by a constant.
FYI : This came up when I am working on computing binomial coefficients with their trignometric form.

Last edited: Aug 28, 2014
2. Aug 29, 2014

### HallsofIvy

Yes, of course, a definite integral, which is, by definition, a constant, can be replaced by that constant.

3. Aug 29, 2014

### PeroK

Those integrals of f(x) must be indefinite integrals. You can't use definite integrals like that in the parts formula.

No, that isn't valid. If you take another step, you'll see that

$$b = a(\left.g(x))\right|_{-∏}^∏-a(\left.g(x))\right|_{-∏}^∏ = 0$$

For all f(x) and g(x). Which clearly isn't correct.

4. Aug 29, 2014

### mathman

Your integration by parts formula looks wrong.
$$\int_{-∏}^∏g'(x)\int_{-∏}^∏ f(x)\,dx \,dx$$
should be:
$$\int_{-∏}^∏g'(x)F(x)dx$$
where F(x) is the indefinite integral of f(x).

5. Aug 31, 2014

### SteliosVas

They must be indefinite integrals in order to do so, unless it is invalid.
$$(\left.g(x)\int_{-∏}^∏ f(x)\,dx)\right|_{-∏}^∏$$
$$(\left.g(x)F(x))\right|_{-∏}^∏$$