Problem with an integral and integration by part

Question

I have a little problem with this question.. :

We have : $$ G(t) = \int_0^t f(e^t)e^{-t} \, dt $$ With f continue in the domain [0,$+\infty$].

and the question is : Prove that

$$\int_0^t G(u) \,du = t\int_0^t e^{-u}f(e^u) \,du \, - \int_0^t ue^{-u}f(e^u) \,du$$

Thank you for yours futur answer

Answer 1

Note that $$ G'(t) = f(e^t)e^{-t}, $$ so that integration by parts gives \begin{align} t\int_0^t f(e^u)e^{-u} \, du - &\int_0^t uf(e^u)e^{-u} \, du = \int_0^t (t-u)G'(u) \, du\\ &= [(t-u)G(u)]_{u=0}^t - \int_0^t (-1) G(u)\,du\\ &= \int_0^t G(u) \, du. \end{align}

Answer 2

Set $H(t)=\int_0^tG(u)\,du$ then $H'(t)=G(t)$ and $H''(t)=f(e^t)e^{-t}$. Now the linear Taylor formula with integral remainder term tells us that $$ H(t)=H(0)+H'(0)t+\int_0^t(t-u)H''(u)\,du, $$ which directly leads to the claimed formula.