Some math jokes (which turn out to be correct)

Prove that $$\int_{0}^{\infty}\frac{1}{(x^2+1)(x^{\pi}+1)}\text{d}x=\int_{0}^{\infty}\frac{1}{(x^2+1)(x^{e}+1)}\text{d}x$$
Prove that $$\int_{0}^{\pi/2}\frac{1}{1+\tan^{\pi}(x)}\text{d}x=\int_{0}^{\pi/2}\frac{1}{1+\tan^{e}(x)}\text{d}x$$

$\textbf{Proof.}$ Both of the integrals are being solved, when the peculiar powers are being replaced by a parameter "a" and then differentiation with respect to "a"  is performed. Actually the integrals are related vie the change of variables $x\to\arctan(x)$.
For the first integral, denote
$$f(x,a)= \frac{1}{(x^2+1)(x^{a}+1)}.$$
Then $$\frac{\partial }{\partial a}f(x,a)=\frac{x^a \log (x)}{\left(x^a+1\right) \left(x^a+1\right)^2}$$
Consider the integral
$$\int_{0}^{1}\frac{\partial }{\partial a}f(x,a)\text{d}x.$$ Making the change $x\to 1/x$ this transforms to the integral
$$\int_{\infty}^{1}\frac{\partial }{\partial a}f\left(\frac{1}{x},a\right)\frac{-1}{x^2}\text{d}x=\int_{1}^{\infty}\frac{\partial }{\partial a}f\left(\frac{1}{x},a\right)\frac{1}{x^2}\text{d}x$$
Now it remains to observe that
$$ \frac{\partial }{\partial a}f\left(\frac{1}{x},a\right)\frac{1}{x^2}=-\frac{\partial }{\partial a}f(x,a)$$
so that
$$\int_{0}^{1}\frac{\partial }{\partial a}f(x,a)\text{d}x=-\int_{1}^{\infty}\frac{\partial }{\partial a}f(x,a)\text{d}x$$
and hence
$$\int_{0}^{\infty}\frac{\partial }{\partial a}f(x,a)\text{d}x=0.$$
Thus  $$\int_{0}^{\infty}f(x,a)\text{d}x$$
is a constant independent of $a$, which proves the equality. Plugging $a=0$ we see moreover that the value of the integrals is $\displaystyle \frac{\pi}{4}$.

$\textbf{Addendum}$. Another solution proceeds as directly making the change of variable $x\to 1/x$. Thus

$$\int_0^\infty\frac{1}{(x^2+1)(x^{a}+1)}\text{d}x=\int_0^\infty\frac{x^a}{(x^2+1)(x^{a}+1)}\text{d}x.$$
Summing both integrals yields the solution.

The second problem could be done in a similar manner.

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