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Evaluate $$\int_{2}^{3}\frac{(x-2)^{100}}{(x-1)}dx$$

Is there an easy method to evaluate the integral. I wanted to try by using long division but I think that it will be very complicate. Any help ,thanks

E.H.E
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    $u=x-1$ and the binomial theorem might make things slightly easier. – JP McCarthy May 12 '15 at 20:04
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    Since the answer is $$\ln 2 -\frac{47979622564155786918478609039662898122617}{69720375229712477164533808935312303556800}$$ I suspect all approaches will be complicated. – mrf May 12 '15 at 20:09
  • The result is $\log (2)-\frac{47979622564155786918478609039662898122617}{69720375229712477164 533808935312303556800}$. I suggest you to not do this by hand. – mickep May 12 '15 at 20:09
  • @mrf The result for the indefinite one according to Wolfram is just some kind of "pattern-like" summation, but still, for the definite one, that really won't help. – Hasan Saad May 12 '15 at 20:11
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    @mrf: the large fraction you cite is $H_{100}-H_{50}$. – robjohn May 12 '15 at 20:35
  • @robjohn Yes. I suspect that the OP hadn't asked the question if he/she knew about harmonic numbers. I interpreted the question as if they wanted a closed formula. – mrf May 12 '15 at 20:38

3 Answers3

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$$ \begin{align} &\int_2^3\frac{(x-2)^{100}}{x-1}\,\mathrm{d}x\\ &=\int_0^1\frac{x^{100}}{x+1}\,\mathrm{d}x\\ &=\int_0^1\left(x^{99}-x^{98}+x^{97}-x^{96}+\dots-1+\frac1{x+1}\right)\,\mathrm{d}x\\ &=\frac1{100}-\frac1{99}+\frac1{98}-\frac1{97}+\dots-1+\log(2)\\ &=2\left(\frac1{100}+\frac1{98}+\dots+\frac12\right) -\left(\frac1{100}+\frac1{99}+\frac1{98}+\dots+1\right) +\log(2)\\[3pt] &=\bbox[5px,border:2px solid #C0A000]{\log(2)+H_{50}-H_{100}} \end{align} $$ where $H_n$ are Harmonic Numbers. Mathematica gives $$ H_{100}-H_{50}=\frac{47979622564155786918478609039662898122617}{69720375229712477164533808935312303556800} $$ which yields the answer mentioned in a couple of comments: $$ \bbox[5px,border:2px solid #C0A000]{\log(2)-\frac{47979622564155786918478609039662898122617}{69720375229712477164533808935312303556800}} $$

The same derivation given above can be easily generalized to give $$ \int_0^1\frac{x^{2k}}{x+1}\,\mathrm{d}x=\log(2)+H_k-H_{2k} $$

robjohn
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The long division takes you $$\frac{(x-2)^{100}}{(x-2)+1}=(x-2)^{99}-(x-2)^{98}+(x-2)^{97}-......+(x-2)-1+\frac{1}{x-1}$$ so $$\int \frac{(x-2)^{100}}{(x-1)}dx=\frac{(x-2)^{100}}{100}-\frac{(x-2)^{99}}{99}+.......-x+\log(x-1)$$ $$\int_{2}^{3}\frac{(x-2)^{100}}{(x-1)}dx=\frac{1}{100}-\frac{1}{99}+\frac{1}{98}.......-1+\log(2)$$ $$=\log(2)-\frac{47979622564155786918478609039662898122617}{69720375229712477164533808935312303556800}$$

E.H.E
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We can easily generalize the development for

$$I_n=\int_2^3 \frac{(x-2)^n}{(x-1)}dx \tag 1$$

We write

$$I_n=\int_2^3 \frac{(x-2)^n}{(x-1)}dx=\int_1^2 \frac{(x-1)^n}{x}dx=\int_1^2 \frac{x-1}{x}(x-1)^{n-1}dx=\frac{1}{n}-I_{n-1}$$

Note that $I_0=\log 2$ and $I_n=\frac{1}{n}-I_{n-1}$. So, we simply iterate to reveal that

$$I_1=-\log 2+1$$

$$I_2=\log 2 -1+\frac12$$

$$I_3=-\log 2+1-\frac12+\frac13$$

$$\begin{align} I_n&=(-1)^n(\log 2-1+1/2-1/3+\cdots +1/n)\\\\ &=(-1)^n\left(\log2 +\sum_{k=1}^{n}\frac{(-1)^k}{k}\right) \tag 2 \end{align}$$

For $n=100$, we have

$$\begin{align} I_{100}&=\log 2+\sum_{k=1}^{100}\frac{(-1)^{k}}{k}\\\\ &=\log 2 -\frac{47979622564155786918478609039662898122617}{69720375229712477164533808935312303556800}\\\\ &\approx. 0.004975 \end{align}$$

where the fraction and the decimal approximation were obtained using Wolfram Alpha.

NOTE:

This approach can be used to prove that the value of the alternating harmonic series is indeed $\log 2$.

Simply observe the limit as $n\to \infty$ of $(1)$ and $(2)$. For $(1)$, we see that $I_n \to 0$ as $n\to \infty$.

We also see that $|I_n| \to |\log 2+\sum_{k=1}^{\infty}\frac{(-1)^n}{n}|$.

Since the term in absolute value must be zero, then $\sum_{k=1}^{\infty}\frac{(-1)^{n+1}}{n}=\log 2$.

Mark Viola
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