To prove a function is integrable over a closed interval

Question

Given that a function $f$ is defined as $$f(x)=1+2x+3x^2+4x^3+...$$. We have to prove that $f$ is continuous on $[0,\frac{1}{8}]$ and evaluate $$\int_{0}^{\frac{1}{8}}f(x)dx$$

I am not sure in a particular area. I have proved $f$ is continuous on $[0,\frac{1}{8}]$. But the evaluation of integral gives

$$\int_{0}^{\frac{1}{8}}f(x)dx=\sum_{n=1}^{\infty}\frac{1}{8^{n}}=\frac{8}{7}(1-({\frac{1}{8})^n})= \frac{8}{7}(1-\frac{1}{8^n})$$.

I think I am wrong somewhere in calculations. Please guide me where I am wrong. Any help or suggestion will be precious.

Answer 1

The first mistake that I was able to detect was that you series should start at $1$, not at $0$:\begin{align}\int_0^{\frac18}f(x)\,\mathrm dx&=\sum_{n=1}^\infty\left(\frac18\right)^n\\&=\frac{\frac18}{1-\frac18}\\&=\frac17.\end{align}Another mistake lies in the fact that the answer should be a number; there should be no $n$ in it.

Answer 2

$f(x)=\displaystyle\sum_{n=0}^{\infty}(n+1)x^{n}$, $x\in[0,1/8]$, then $f(x)\leq\displaystyle\sum_{n=0}^{\infty}(n+1)/8^{n}<\infty$ by ratio test, and by Weierstrass M-Test, the series is uniformly convergent, and we are allowed to swipe the sum and integral: $\displaystyle\int_{0}^{1/8}f(x)dx=\displaystyle\sum_{n=0}^{\infty}(n+1)\int_{0}^{1/8}x^{n}dx=\sum_{n=0}^{\infty}(1/8)^{n+1}=\dfrac{(1/8)}{1-(1/8)}=1/7$.

The mistake there is that there should be no $n$ left, there is the limit taken.

Answer 3

As an alternative approach, it is not difficult to notice that $(1-x)^2 f(x)$ has a simple expression.
In general, if $f(x)=\sum_{n\geq 0}p(n) x^n$ with $p$ being a polynomial with degree $d$, $(1-x)^{d+1}f(x)$ has a simple expression, since the multiplication by $(1-x)$ acts on the coefficients of $f(x)$ as the backward difference operator. In our case $$(1-x)^2\sum_{n\geq 0}(n+1)x^n =1$$ (massive cancellation!) for any $x\in(-1,1)$, hence you are just asked to check that $\frac{1}{(1-x)^2}$ is continuous on $\left[0,\frac{1}{8}\right]$, which is obvious since $(1-x)^2$ is continuous and bounded below by a positive constant, and to compute $$ \int_{0}^{1/8}\frac{dx}{(1-x)^2} = \int_{7/8}^{1}\frac{dx}{x^2}=\frac{8}{7}-1=\frac{1}{7}.$$ Of course, the RHS can also be computed from $$ \int_{0}^{+\infty}\sum_{n\geq 0}(n+1) x^n\,dx = \sum_{n\geq 0}(n+1)\int_{0}^{1/8}x^n\,dx = \sum_{n\geq 0}\frac{1}{8^{n+1}} = \frac{1}{7}\quad\text{(geometric series).}$$

Answer 4

Another approach: Evaluate the series first, then integrate $$ f(x)=\sum_{n=0}^\infty (n+1)x^n=\sum_{n=0}^\infty nx^n+ \sum_{n=0}^\infty x^n\\ =x\sum_{n=0}^\infty nx^{n-1}+\frac{1}{1-x}\\ =x\frac{1}{(1-x)^2}+\frac{1}{1-x} $$ Now both integrating and checking continuity (it's the sum of two rational functions, neither singularity is contained in the interval $[0,1/8]$) isn't too bad. Indeed, $$ \int f(x)\mathrm dx=\int \frac{x}{(1-x)^2}\mathrm dx-\ln|1-x|\\ \stackrel{u=1-x}=-\ln|1-x|-\int\frac{1}{u^2}\mathrm du+\int\frac1u\mathrm du\\ =\frac{1}{1-x}+C $$ Now plugging in we find $$ \int_0^{1/8}f(x)\mathrm dx=\frac{1}{1-1/8}-1=1/7 $$