- #1
epr2008
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I'm 19 and i just finished ode at a community college. there are no more math classes for me to take there so I've been planning teaching myself this semester. I was good at calc and ode's so i read about integral equations and figured i would start. I am using a book by Peter Collins but he's not very good at explaining the in between steps. he always says "i'll leave this for the reader to prove." well some of these are really not easy to see... so, I started off trying to prove a lemma he introduced in the first chapter and out of nowhere i ended up with something that i think is much more enlightening i just don't know if its right, so i was looking for some input...
The lemma to be proved:
Suppose that f:[a,b][tex]\rightarrow[/tex][tex]\Re[/tex] is continuous. Then
[tex]\int\limits_a^x {\int\limits_a^{x'} {f(t)dtdx' = \int\limits_a^x {(x - t)f(t)dt} } } [/tex]
Ok so at first before i figured out what he was talking about i started out like the book did with the rhs of the equation. and i thought about it and i said well what is the integral of that? ok integration by parts...
[tex]\begin{array}{l}
F(x) = \int\limits_a^x {(x - t)f(t)dt} \\
u = f(t) \\
du = f'(t)dt \\
dv = (x - t)dt \\
v = xt - {\textstyle{1 \over 2}}{t^2} \\
F(x) = (xt - {\textstyle{1 \over 2}}{t^2})f(t) - \int {(xt - {\textstyle{1 \over 2}}{t^2})f'(t)dt} ]\nolimits_a^x \\
\end{array}[/tex]
Now if f is a continuously differentiable function in t then integration by parts will eventually yield a differential equation with variable coefficients of the form
[tex]F(x) = \sum\limits_0^\infty {{{( - 1)}^n}} {f^n}(t)\frac{{(x + t)n + 2x + t}}{{(n + 1)(n + 2)}}{t^{n + 1}} ]\nolimits_a^x [/tex]
Now obviously had i chosen u=(x-t) and dv=f(t)dt then i would have come up with another integral but with no derivatives in the equation right? Is this somewhat along the lines of the basic theory of differential and integral equations or am i just wrong?
The lemma to be proved:
Suppose that f:[a,b][tex]\rightarrow[/tex][tex]\Re[/tex] is continuous. Then
[tex]\int\limits_a^x {\int\limits_a^{x'} {f(t)dtdx' = \int\limits_a^x {(x - t)f(t)dt} } } [/tex]
Ok so at first before i figured out what he was talking about i started out like the book did with the rhs of the equation. and i thought about it and i said well what is the integral of that? ok integration by parts...
[tex]\begin{array}{l}
F(x) = \int\limits_a^x {(x - t)f(t)dt} \\
u = f(t) \\
du = f'(t)dt \\
dv = (x - t)dt \\
v = xt - {\textstyle{1 \over 2}}{t^2} \\
F(x) = (xt - {\textstyle{1 \over 2}}{t^2})f(t) - \int {(xt - {\textstyle{1 \over 2}}{t^2})f'(t)dt} ]\nolimits_a^x \\
\end{array}[/tex]
Now if f is a continuously differentiable function in t then integration by parts will eventually yield a differential equation with variable coefficients of the form
[tex]F(x) = \sum\limits_0^\infty {{{( - 1)}^n}} {f^n}(t)\frac{{(x + t)n + 2x + t}}{{(n + 1)(n + 2)}}{t^{n + 1}} ]\nolimits_a^x [/tex]
Now obviously had i chosen u=(x-t) and dv=f(t)dt then i would have come up with another integral but with no derivatives in the equation right? Is this somewhat along the lines of the basic theory of differential and integral equations or am i just wrong?