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For example, Let us consider a function, say,

$f(x)=\frac{1}{1+tan^{\sqrt 2}x}$

This cannot be computed when it is indefinite. But, this can be integrated when we try to compute it with definite integration, using symmetry of this function.

Please refer this link for the graph and some explanation.

So my question now is, How do we say whether the graph of a given function is symmetrical or not. Is there any specific conditions under which the graph would be symmetrical?

Thanks in advance.

This is not a duplicate of the $\int\frac{1}{1+tan^{\alpha}x}dx$ question. This question is just used as an example.

Mittens
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Fetray
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3 Answers3

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Here, if I set $f(x) = \frac1{1+\tan^{\sqrt2}(x+\frac\pi4)} - \frac12$, then $f(x) = -f(-x)$.

As an application, this makes computing the integral of your function very easy:

$$ \int_0^{\pi/2} \frac{\text dx}{1+\tan^{\sqrt2}(x)} = \int_0^{\pi/2} \left(f\left(x-\frac\pi4\right)+\frac12\right)\text dx = \int_{-\pi/4}^{\pi/4} f(x)\text dx +\frac\pi4 = \frac\pi4. $$

These operations are key to define what a symmetry is

  • $f(x) \mapsto f(x-a)$ shifts the graph of $f$ to the right ($a > 0$) or to the left ($a < 0$)
  • $f(x) \mapsto f(x) + a$ shifts the graph of $f$ upward ($a > 0$) or downward ($a < 0$)
  • $f(x) \mapsto f(-x) $ flips the graph of $f$ along the $y$ axis
  • $f(x) \mapsto -f(x) $ flips the graph of $f$ along the $x$ axis

So if a combination of these operations on $f$ equals another combination, then we can say we have a symmetry.

For example, if for some $a, b$, $g(a-x) - b = - g(a+x) + b$ this constitutes a symmetry and $f(x) = g(a-x) - b$ is an odd function. So more generally than before

$$ \int_0^{2a} g(x) \text dx = 2ab. $$

Nolord
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In your case,

$$\mathcal {f} \left( x \right) + \mathcal {f} \left( \textstyle \frac {\pi}{2} - x \right) = 1.$$

Let

$$\mathcal {f} \left( x \right) = \mathcal {g} \left( x \right) + \textstyle \frac {1}{2}.$$

Then

$$\mathcal {g} \left( x \right) + \mathcal {g} \left( \textstyle \frac {\pi}{2} - x \right) = 0.$$

or equivalently,

$$\mathcal {g} \left( x + \textstyle \frac {\pi}{4} \right) + \mathcal {g} \left( \textstyle \frac {\pi}{4} - x \right) = 0.$$

(Compare this to the defining equation of oddity.)

enter image description here

Simon
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-1

The function is defined as

$$\frac{1}{1+\tan^{\sqrt 2}(x)}= \frac{1}{1+e^{\sqrt 2} \log (\tan x)}$$. The $\log$ is multivalued. The restricted represenatation in the complex plane has a cut on the negative real line. So as a real integral its only defined for positive interval of $x\mapsto \tan x$, e.g. $x \in (0,\pi)$

Using the standard substitution $\tan(\frac{x}{2})=y$ yields

$$\int \frac{1}{1+\left(\tan(\frac{x}{2})\right)^\alpha} dx \propto \int \frac{dy}{(1+y^2)(1+y^\alpha)} , \quad y \ge 0$$

Real, noninteger powers of negative numbers are undefined.

Roland F
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