Can integral be replaced by constant

[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.

 

[HallsofIvy]

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

 

[PeroK]

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 = $$(\left.g(x)\int_{-∏}^∏ f(x)\,dx)\right|_{-∏}^∏-\int_{-∏}^∏g'(x)\int_{-∏}^∏ f(x)\,dx \,dx$$

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

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

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.

 

[mathman]

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.

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).

 

[SteliosVas]

They must be indefinite integrals in order to do so, unless it is invalid.
I think you have made a mistake somewhere in your working.

 

[mathman]

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.

Further error:
$$(\left.g(x)\int_{-∏}^∏ f(x)\,dx)\right|_{-∏}^∏$$
should be:
$$(\left.g(x)F(x))\right|_{-∏}^∏$$
where F(x) is the indefinte integral of f(x)