- #1
smslce
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Let a,b are constants.
[tex]a=\int_{-∏}^∏ f(x)\,dx [/tex]
and
[tex]b=\int_{-∏}^∏ {g(x)f(x)}\,dx,[/tex]
Then by integration byparts, we have ,
b = [tex](\left.g(x)\int_{-∏}^∏ f(x)\,dx)\right|_{-∏}^∏-\int_{-∏}^∏g'(x)\int_{-∏}^∏ f(x)\,dx \,dx [/tex]
Is below step valid.
b = [tex]a(\left.g(x))\right|_{-∏}^∏-a\int_{-∏}^∏g'(x)\,dx [/tex]
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.
[tex]a=\int_{-∏}^∏ f(x)\,dx [/tex]
and
[tex]b=\int_{-∏}^∏ {g(x)f(x)}\,dx,[/tex]
Then by integration byparts, we have ,
b = [tex](\left.g(x)\int_{-∏}^∏ f(x)\,dx)\right|_{-∏}^∏-\int_{-∏}^∏g'(x)\int_{-∏}^∏ f(x)\,dx \,dx [/tex]
Is below step valid.
b = [tex]a(\left.g(x))\right|_{-∏}^∏-a\int_{-∏}^∏g'(x)\,dx [/tex]
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.
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