I recently noticed that this really weird equation actually carries a closed form!

$$\int_0^1 \left(\frac{x^x}{(1-x)^{1-x}}-\frac{(1-x)^{1-x}}{x^x}\right)\text{d}x=0$$

I honestly do not know how to prove this amazing result! I do not know nearly enough about the sophomore's dream integral properties to answer this question, which I have been trying to apply here. (If possible, please stay with real methods, as I do not know contour integration yet)


It is important to note some symmetry:

Consider: $$I=\int_0^1 \frac{x^x}{(1-x)^{1-x}}\text{d}x$$

If you substitute $x\to 1-x, \text{d}x\to -\text{d}x$

You are left with:

$$I=-\int_{1-0}^{1-1}\frac{(1-x)^{(1-x)}}{(1-(1-x))^{(1-(1-x))}}\text{d}x=\int_0^1 \frac{(1-x)^{1-x}}{x^x}\text{d}x$$

And your integral is only $I-I=0$, your result. You didn't even need to know $x^x$ properties!

In fact, this can be further generalized:

$$\int_0^1 \frac{f(x)}{f(1-x)}\text{d}x=\int_0^1 \frac{f(1-x)}{f(x)}\text{d}x$$

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    $\begingroup$ I did not expect such a positive reaction from this answer! Thank you all! $\endgroup$ – user331275 Apr 15 '16 at 19:51
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    $\begingroup$ You can actually take this even further: $\int_0^n f(x)\text{d}x=\int_0^n f(n-x)\text{d}x$! $\endgroup$ – Pockets Apr 15 '16 at 20:51
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    $\begingroup$ It's actually rather surprising that you didn't notice what Pocket pointed out haha.. It's just area under the curve reflected across a line of symmetry. $\endgroup$ – user21820 Apr 16 '16 at 7:51
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    $\begingroup$ I believe that you guys are misinterpreting what I am getting at. When I did my proof, I just didn't see it. However, now that you show it to me, it does make sense :) $\endgroup$ – user331275 Apr 16 '16 at 23:41
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    $\begingroup$ +1, but it should be noted that this only proves that if the integral exists, then it must be zero. $\endgroup$ – ruakh Apr 17 '16 at 0:32

So you have an integral: $$I = \int_0^1 \left(\frac{x^x}{(1-x)^{1-x}}-\frac{(1-x)^{1-x}}{x^x}\right)\text{d}x$$

Let's split the area of integration by half:

$$I = \int_0^{\tfrac 12} \left(\frac{x^x}{(1-x)^{1-x}}-\frac{(1-x)^{1-x}}{x^x}\right)\text{d}x + \int_{\tfrac 12}^1 \left(\frac{x^x}{(1-x)^{1-x}}-\frac{(1-x)^{1-x}}{x^x}\right)\text{d}x $$

and substitute $y=1-x$ in the second term:

$$I = \int_0^{\tfrac 12} \left(\frac{x^x}{(1-x)^{1-x}}-\frac{(1-x)^{1-x}}{x^x}\right)\text{d}x + \int_{\tfrac 12}^0 \left(\frac{(1-y)^{1-y}}{y^y}-\frac{y^y}{(1-y)^{1-y}}\right)(-\text{d}y) $$ $$ =\int_0^{\tfrac 12} \left(\frac{x^x}{(1-x)^{1-x}}-\frac{(1-x)^{1-x}}{x^x}\right)\text{d}x + \int_{\tfrac 12}^0 \left(\frac{y^y}{(1-y)^{1-y}}-\frac{(1-y)^{1-y}}{y^y}\right)\text{d}y $$

Now we swap the limits of integration for $y$: $$I = \int_0^{\tfrac 12} \left(\frac{x^x}{(1-x)^{1-x}}-\frac{(1-x)^{1-x}}{x^x}\right)\text{d}x - \int_0^{\tfrac 12} \left(\frac{y^y}{(1-y)^{1-y}}-\frac{(1-y)^{1-y}}{y^y}\right)\text{d}y $$ and after a few minutes of staring up we see the expression is $$I = K - K$$ with $$K = \int_0^{\tfrac 12} \left(\frac{x^x}{(1-x)^{1-x}}-\frac{(1-x)^{1-x}}{x^x}\right)\text{d}x$$

Hence $I = 0$.


As property of definite integrals it is known that for any function integrand $f(x)$

$$\int_0^a{f(x)}\text{d}x = \int_0^a{f(a-x)}\text{d}x$$

Geometrically this means the area under the curve when flipped about the line $ x= \dfrac{a}{2}$ as a rigid figure cannot change. The property can be stated symbolically as in one particular generalisation where ${p} $ is a constant:

$$\int_0^1 \frac{f(x)^p}{f(1-x)^p}\text{d}x=\int_0^1 \frac{f(1-x)^p}{f(x)^p}\text{d}x.$$

For p = 1 you can write the integrand also as a product in another example.

$$ \int_0^{\pi}{\cos^5 \phi \sin \phi }\, d \phi = \int_0^{\pi}{\sin ^5 \phi \cos \phi }\, d \phi $$

I am sure the generalisation takes a fanciful attention.Earlier days I would imagine a "curved roof trapezoid " and flip it without change of area under the roof.


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