I recently came across a very nice wordpress page. There are a plenty of nice math in that page that I thoroughly enjoyed, especially the Distances to a line from the vertices of a regular polygon. In this post, we look at one of the results based on Integrals not being dependent on a parameter from the same wordpress page.
The author shows that
$\displaystyle \int_0^\infty f(x^2)\,dx=\int_0^\infty f((x-a/x)^2)\,dx$, $\text{Re}(a)\leq0$
This result can be further generalized. In fact, the presence of the square term in the facilitates the proof of the result via differential under the integral sign (sadly, only via that route), but that doesn't mean we can't take a different route to the general result. Taking a cue from Problem B4 of 29th Putnam 1968, we show that,
$\displaystyle \int_{-\infty}^\infty f(x)\,dx=\int_{-\infty}^\infty f(x-a/x)\,dx$ provided one of them exists.
For example, here are the WolframAlpha results for $\sin x / x$ and $\sin(x-1/x)/(x-1/x)$. It's surprising that while Mathematica calculates the first case exactly as $\pi$ in the blink of an eye, it fails to evaluate the second.
Let $I=\displaystyle \int_{-\infty}^\infty f(x)\,dx$
Substitute $x=y-a/y$. Now comes the cute part. Note that when $y=0$, $x=-\infty$ and when $y=\infty$, $x=\infty$. Also, $dx=(1+\frac{a}{y^2})dy$
Therefore,
$\begin{align}
I&=\displaystyle \int_0^\infty f(y-a/y)\left(1+\frac{a}{y^2}\right)\,dy\\
&=\displaystyle \int_0^\infty f(y-a/y)\,dy+\displaystyle \int_0^\infty f(y-a/y)\frac{a}{y^2}\,dy\\
\end{align}$
Now, we make the substitution, $t=-\frac{a}{y}$ in the second integral. Therefore, $dt=\frac{a}{y^2}dy$. Also, $y=0,t=-\infty$ and $y=\infty,t=0$. Hence,
$\begin{align}
I&=\displaystyle \int_0^\infty f(y-a/y)\,dy+\displaystyle \int_{-\infty}^0 f(-a/t+t)\,dt\\
&=\displaystyle \int_{-\infty}^\infty f(x-a/x)\,dx
\end{align}$
which proves the result.
In any case, just like the author of the wordpress page, I too was surprised by the fact that the transformation $x \to x-a/x$ changes the shape of the curve drastically keeping the area constant. A non-calculus based proof of this result would be a feast to the senses I guess but I couldn't find it yet. See ya again with a nice post.
Until then
Yours Aye
Me
The author shows that
$\displaystyle \int_0^\infty f(x^2)\,dx=\int_0^\infty f((x-a/x)^2)\,dx$, $\text{Re}(a)\leq0$
This result can be further generalized. In fact, the presence of the square term in the facilitates the proof of the result via differential under the integral sign (sadly, only via that route), but that doesn't mean we can't take a different route to the general result. Taking a cue from Problem B4 of 29th Putnam 1968, we show that,
$\displaystyle \int_{-\infty}^\infty f(x)\,dx=\int_{-\infty}^\infty f(x-a/x)\,dx$ provided one of them exists.
For example, here are the WolframAlpha results for $\sin x / x$ and $\sin(x-1/x)/(x-1/x)$. It's surprising that while Mathematica calculates the first case exactly as $\pi$ in the blink of an eye, it fails to evaluate the second.
Let $I=\displaystyle \int_{-\infty}^\infty f(x)\,dx$
Substitute $x=y-a/y$. Now comes the cute part. Note that when $y=0$, $x=-\infty$ and when $y=\infty$, $x=\infty$. Also, $dx=(1+\frac{a}{y^2})dy$
Therefore,
$\begin{align}
I&=\displaystyle \int_0^\infty f(y-a/y)\left(1+\frac{a}{y^2}\right)\,dy\\
&=\displaystyle \int_0^\infty f(y-a/y)\,dy+\displaystyle \int_0^\infty f(y-a/y)\frac{a}{y^2}\,dy\\
\end{align}$
Now, we make the substitution, $t=-\frac{a}{y}$ in the second integral. Therefore, $dt=\frac{a}{y^2}dy$. Also, $y=0,t=-\infty$ and $y=\infty,t=0$. Hence,
$\begin{align}
I&=\displaystyle \int_0^\infty f(y-a/y)\,dy+\displaystyle \int_{-\infty}^0 f(-a/t+t)\,dt\\
&=\displaystyle \int_{-\infty}^\infty f(x-a/x)\,dx
\end{align}$
which proves the result.
In any case, just like the author of the wordpress page, I too was surprised by the fact that the transformation $x \to x-a/x$ changes the shape of the curve drastically keeping the area constant. A non-calculus based proof of this result would be a feast to the senses I guess but I couldn't find it yet. See ya again with a nice post.
Until then
Yours Aye
Me
