Question

When can I take summation out of integral? Please, help.
\(\int_a^b \sum_{i\geq 1} f(x)dx\)

Answer 1

@satellite73

Answer 2

@mathstudent55

Answer 3

i believe you can just replace them, if you know what i'm saying

Answer 4

My particular problem:
\[int_0^1\int_0^1\sum_{i\geq 1} (xy)^i dxdy\]

Answer 5

@ParthKohli it is NOT replace or not, it is integration of a sum.

Answer 6

the integral of a sum is also a sum of integrals, if that makes your problem easier to solve.

Answer 7

I believe that we have some constraint for f(x), but not sure what they are

Answer 8

continuity is for sure, without sum, to take integration of a function, it must be continuous on the interval, but it is so weak in this case

Answer 9

yeah, I agree with you. in this case, I'm not sure, but it does seem that we can apply that rule. otherwise, I'm only speaking from my experience of single variable calculus.

Answer 10

In this case, if there were no series.....

\[\int_0^1\int_0^1 dxdy \qquad (xy)^i\]
\[= \int_0^1 dy \qquad \left[ \dfrac{x^{i+1} y^ i}{i + 1} \right]_0^1 \]
\[= \left[ \dfrac{y^{i+1} }{(i + 1)^2} \right]_0^1\]
\[= \dfrac{1 }{(i + 1)^2}\]

So:

\[\int_0^1\int_0^1 dxdy \qquad \sum_{i\geq 1} (xy)^i\]
\[= \int_0^1\int_0^1 dxdy \qquad xy + (xy)^2 + (xy)^3 + \dots\]

\[ =\sum_{i\geq 1} \dfrac{1 }{(i + 1)^2}\]

\[= \sum_{i\geq 1} ~ \int_0^1\int_0^1 dxdy \qquad \]

et voila! i think.


but if you repeat this exercise, say, inside the triangle under y = x, x = 0 to x = 1, then first, ignoring the series,....

\[\int\limits_{y=0}^1~~\int\limits_{x=y}^1 dxdy \qquad (xy)^i\]
\[= \int\limits_0^1 dy \qquad \left[ \dfrac{x^{i+1} y^ i}{i + 1} \right]_{x=y}^1 \]
\[= \int\limits_0^1 dy \qquad \dfrac{ y^ {i}}{i + 1} - \dfrac{ y^ {2i+1}}{i + 1} \]
\[= \qquad \left[ \dfrac{ y^ {i+1}}{(i + 1)^2} - \dfrac{ y^ {2(i+1)}}{2(i + 1)^2} \right]_0^1 \]

\[= \dfrac{1}{2(1+i)^2}\]

so you're kinda integrating a constant again. which means you can lift it outside.

that seems to wok but i remain vert sceptical. double integrals are iterated integrals so it doesn't make sense that you can make it all so simple.

that probably doesn't help either, lol!!!

Answer 11

LOLOLLOL.... anyway, thanks for the help. I think I got it.