I am trying to prove that $$\begin{equation}\int_x^{x+1}\left(\int_0^{v} (u-0)f(u)\textrm{d}u+\int_v^{1} (u-1)f(u)\textrm{d}u\right)\textrm{d}v=\\\int_0^x\int_v^{v+1}f(u)\textrm{d}u\textrm{d}v\end{equation}.$$

With the following sympy code i verified it symbolically for a few cases.

from sympy import *

f=lambda x:exp(x);
print expand(integrate(integrate((u-0)*f(u), 
print expand(integrate(integrate(f(u),(u,v,v+1)),(v,0,x)))

Call $F$ the antiderivative of $f$ such that $F(1)=0$.

Call $\frak{F}$ the antiderivative of $F$ such that $\frak{F}$$(1)=0$.

Now show that both expressions give the same result, i.e., :

$${\frak F}(x+1)-{\frak F}(x)+{\frak F}(0).$$

Method : Write your LHS integral under the form :

$$\begin{equation}\int_x^{x+1}\left(\int_0^1 uf(u)\textrm{d}u-\int_v^{1}f(u)\textrm{d}u\right)\textrm{d}v\end{equation}.$$

Then integrate inside the large parentheses by parts in the first expression, giving

$$\int_{x}^{x+1}([uF(u)]_0^1-\int_0^1 F(u)du + F(v)-F(1)$$

Taking into account $F(1)=0$ and ${\frak F}(1)=0$ :

$$=\int_{x}^{x+1}({\frak F}(0) + F(v))dv$$

I let you find the rest...

  • $\begingroup$ $\int_0^{v} (u-0)f(u)\textrm{d}u+\int_v^{1} (u-1)f(u)\textrm{d}u=\int_0^{v} uf(u)\textrm{d}u+\int_v^{1} uf(u)\textrm{d}u-\int_v^{1}f(u)\textrm{d}u$ $\endgroup$ – Peter Sheldrick Dec 10 '18 at 10:10
  • $\begingroup$ You are right. Corrected. $\endgroup$ – Jean Marie Dec 10 '18 at 10:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.