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Evaluate $$\int_{0}^{1} \dfrac{1}{1 + \left(1 - \dfrac{1}{x}\right)^{2015}} dx$$

Wolfram says it's 1/2 but I thought that you can't integrate this because there is a discontinuity at x = 0.

Progress: The inverse substitution $x$ = $\dfrac{1}{u+1}$ gave me this: http://www.wolframalpha.com/input/?i=integrate+1%2F((u%2B1)%5E2(1-u%5E2015))+from+0+to+infinity

user1952500
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2 Answers2

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Hint:If you simplify the function ,you will see $x=0$ is not a big problem $$\int_{0}^{1}\frac{1}{1+(1-\frac1x)^{2015}}=\\ \int_{0}^{1}\frac{1}{1+(\frac{x-1}{x})^{2015}}=\\ \int_{0}^{1}\frac{1}{1+(\frac{(x-1)^{2015}}{x^{2015}})}=\\ \int_{0}^{1}\frac{1}{\frac{x^{2015}+(x-1)^{2015}}{x^{2015}}}=\\ \int_{0}^{1}\frac{x^{2015}}{x^{2015}+(x-1)^{2015}}dx$$

This well known $\int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx\\\\$

$$I=\int_{0}^{1}\frac{x^{2015}}{x^{2015}+(x-1)^{2015}}dx=\\ \int_{0}^{1}\frac{(1-x)^{2015}}{(1-x)^{2015}+(1-x-1)^{2015}}dx=\\\int_{0}^{1}\frac{(1-x)^{2015}}{(1-x)^{2015}+(-x)^{2015}}dx=\\ \int_{0}^{1}\frac{-(x-1)^{2015}}{-(x-1)^{2015}-(x)^{2015}}dx=\\\int_{0}^{1}\frac{(x-1)^{2015}}{(x-1)^{2015}+(x)^{2015}}dx=\\$$ now $$I+I=\int_{0}^{1}\frac{(x-1)^{2015}}{(x-1)^{2015}+(x)^{2015}}dx+\int_{0}^{1}\frac{(x)^{2015}}{(x-1)^{2015}+(x)^{2015}}dx=\\\int_{0}^{1}\frac{(x-1)^{2015}+(x)^{2015}}{(x-1)^{2015}+(x)^{2015}}dx=\\\int_{0}^{1}1dx=1$$so $$2I=1 \to I =\frac12$$

Ayush Prakash
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Khosrotash
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    There is a little problem: All your denominators vanish at $x=1/2$. The integrals make sense in the principal value sense around this point, but not otherwise. I gave some details in another answer. – Joonas Ilmavirta Jul 05 '17 at 08:10
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The Riemann integral does not exist at all!

Let's denote $f(x)=\frac{1}{1 + \left(1 - 1/x\right)^{2015}}$. This function is not actually singular at $x=0$, but in fact $\lim_{x\to0}f(x)=0$. If you define $f$ by the above formula for $x\neq0$ and $f(0)=0$, the function is indeed continuous at zero. Redefining the value at a single point has no effect on the integral.

However, the function is ill-behaved at $x=\frac12$. To see what the singularity looks like, you can do it by hand or ask WolframAlpha to do a series expansion at that point. The singular part looks like $8060/(x-1/2)$. This in fact implies that the function is not integrable: the integral $\int_0^1f(x)dx$ does not exist as a Riemann integral.

However, the principal value integral $$ pv\int_0^1f(x)dx = \lim_{h\to0} \left( \int_0^{1/2-h}f(x)dx + \int_{1/2+h}^1f(x)dx \right) $$ does exist and the value is indeed $\frac12$ as explained in Khosrotash's answer.

Unless a principal value integral is meant, the integral does not exist. That is, $\int_0^1f(x)dx=\frac12$ in the same sense as $\int_{-1}^1x^{-1}dx=0$.

Why should one expect trouble at $x=1/2$ but not at $x=0$, then? As $x\to0$, we have indeed $1/x\to\infty$, and so the denominator of $f(x)$ goes to infinity and thus $f(x)$ goes to zero. This is not an issue. Issues arise if you need to divide by zero. And indeed $1 + \left(1 - 1/x\right)^{2015}=0$ on the interval of interest if and only if $x=1/2$. Hence, the integral does not exist.

Community
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Joonas Ilmavirta
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