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Consider

$$\int_{0}^{a}\left(\sum_{j=0}^{n-1}(-x)^j\right)^{b-1}\cdot\sum_{j=0}^{n-1}(jb+1)(-x)^j\mathrm dx=I\tag1$$ How does one show that $$I=a(1-a+a^2-a^3+\cdots+a^{n-1})^{b}$$ $(a,b)>0$

expanded

$$\int_{0}^{a}(1-x+x^2-\cdots-x^{n-1})^{b-1}(1-(b+1)x+(2b+1)x^2-\cdots[(n-1)b+1]x^{n-1})\mathrm dx=I$$

Assume $a<1$, then

$$\int_{0}^{a}(1+x)^{1-b}\cdot\sum_{j=0}^{\infty}(jb+1)(-x)^j\mathrm dx\tag2$$

But doesn't really help to prove $(1)$

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    Where the crap do you keep coming up with these problems 0_0 I mean, they are fun to solve, but you seem to have had an infinite supply recently – 2017-02-14
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    Go to work daily and ideas just happened to pop up while travelllling on the bus and train, so I wrote it down and start to work on it. Putting these ideas and formulas into my book and keep in the cupboard is quite dull and useless so might as well write it here. – 2017-02-14
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    Don't you get maths ideas come to you while travelling around @Brevan Ellefsen? – 2017-02-14
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    evidently not as much as yourself XD I'm bad at doing mental math, and I'm unable to keep track of symbols in my head very well. As such, I tend to invent most of my problems when I'm sitting down. When commuting somewhere I tend to think about problems I've already formulated. – 2017-02-14
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    Also, first step I would take on this problem is to take $-x \to x$. I fail to see why you chose to use $-x$ instead... – 2017-02-14
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    The first series is equal to $\frac{x^n - 1}{x-1}$ after the transformation – 2017-02-14
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    I have look through Wikipedia or wolfram to find the closed form for theses sums, but couldn't find any. – 2017-02-14
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    first one is a geometric series, second one is a modified geometric series (multiply the terms out and you find that it's a geometric series plus the derivative of a geometric series) – 2017-02-14
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    because we have closed forms for the geometric-like series in your post I'm fairly certain we can get a closed form... Biggest problem is definitely just the integral. How hard the function is to integrate will depend on what we get when we multiply the closed forms of the sums – 2017-02-14
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    my biggest fear is that we might get a quartic polynomial in the denominator, which are a beast to integrate (though always possible, at least when the numerator is a polynomial) – 2017-02-14

2 Answers 2

3

Define $f(x) = \sum_{j=0}^{n-1}(-x)^j$ then the integral can be written $$I(a) = \int_{0}^{a}f(x)^{b} + x\frac{d}{dx}f(x)^{b}{\rm d}x = \int_0^a\frac{d}{dx}\left[xf(x)^b\right]{\rm d}x = af(a)^b$$

2

Winther's answer is definitely the shortest way. Here I will detail out how we get there, and show what a solution would be like if were to actually try differentiation. We first run the substitution $-x \to x$ (not necessary at all, but it makes the geometric sum more obvious for beginners in my opinion) $$\int_{0}^{a}\left(\sum_{j=0}^{n-1}x^j\right)^{b-1}\left(\sum_{j=0}^{n-1}(jb+1)x^j\right)\mathrm dx$$

We first note that, by the geometric series formula, $$\sum_{j=0}^{n-1}x^j = \frac{x^n -1}{x-1}\tag{1}$$ We can also split the second sum into the form $$b\sum_{j=0}^{n-1}jx^j+\sum_{j=0}^{n-1}x^j $$ The second sum is clearly the geometric series we had before; the first can be written as $$b\sum_{j=0}^{n-1}jx^j=bx\sum_{j=0}^{n-1}\frac{d}{dx}\left(x^j\right)=bx\frac{d}{dx}\sum_{j=0}^{n-1}x^j \tag{2}$$ We can use $(1)$ and $(2)$ we can write the original problem as $$\int_{0}^{a}\left(\frac{x^n -1}{x-1}\right)^{b-1}\left(bx\frac{d}{dx}\left(\frac{x^n -1}{x-1}\right)+\frac{x^n -1}{x-1}\right)\mathrm dx \tag{3}$$ If we try to actually differentiate we get $$\int_{0}^{a}\frac{\left(\frac{x^n-1}{x-1}\right)^{b-1} \left(x^n ((bn-b+1) x-b n-1)+(b-1) x+1\right)}{(x-1)^2}\mathrm dx$$
This could be simplified, if you had the patience of a god. I will leave that up to you though, and just point out that we can write $(3)$ in the form $$\int_{0}^{a}f(x)^{b-1}\left(bx\frac{d}{dx}f(x)+f(x)\right)\mathrm dx$$ $$=\int_{0}^{a} \left(bx\frac{d}{dx}f(x)f(x)^{b-1}+f(x)^b\right)\mathrm dx$$ $$=\int_{0}^{a} \bigg(bxf'(x)f(x)^{b-1}+f(x)^b\bigg)\mathrm dx$$ Applying the chain rule in reverse we find that $x\frac{d}{dx}f(x)^b = f(x)^{b-1}\cdot bxf'(x)$, and we apply this to gett $$=\int_{0}^{a} \bigg(x\frac{d}{dx}f(x)^b+f(x)^b \bigg)\mathrm dx$$
Which leads us into Winther's answer