How may I calculate Integration by parts for the following multivariable function?

$$I_n(x)=\int_{a}^{x} (x-t)^{n}f(t)dt$$


1 Answer 1


I can write out the formula for you, I guess, assuming that $F' =f$ exists: $$ \int_a^x (x-t)^n f(t)dt = \int_a^x (x-t)^n dF(t)t = [(x-t)^n F(t)]_a^x + \int_a^x n(x-t)^{n-1} F(t) dt $$ so that $$ \int_a^x (x-t)^n f(t)dt = -(x-a)^n F(a) + \int_a^x n(x-t)^{n-1} F(t) dt $$ Is that what you are looking for? Does $f$ have any other properties, like those of a CDF, for example?

  • $\begingroup$ Thanks, is it possible to replace ∫𝑛(𝑥−𝑡)𝑛−1𝐹(𝑡)𝑑𝑡 with something like $I_n$ or $I_{n+1}$ (with some edits of course). I know that f(t) is continuous $\endgroup$
    – Daniel98
    Jun 19, 2020 at 15:05

Not the answer you're looking for? Browse other questions tagged .