GeoDom
miércoles, 17 de julio de 2024
Showing $\frac{1}{e^{i\alpha_2}} + \frac{1}{e^{i\beta_2}} + \frac{1}{e^{i\gamma_2}} = \frac{1}{e^{i(\alpha_2+\beta_2+\gamma_2)}}$
viernes, 5 de julio de 2024
Integrals of the form $\int f\left(x,\frac{\sqrt{x+m}}{\sqrt{x+n}}\right)\,dx$
Let $b-a=m$ and $b+a=n$, where $a$ and $b$ are real numbers. Then the following identities hold:
$$\frac{1-e^{\pm\text{i}\alpha}}{1+e^{\pm\text{i}\alpha}}=\tan{\left(\frac12\sec^{-1}\left(\frac{x+b}{a}\right)\right)}=\frac{\sqrt{x+b-a}}{\sqrt{x+b+a}}=\frac{\sqrt{x+m}}{\sqrt{x+n}},\tag{1}$$
$$e^{\pm\text{i}\alpha}=\tan\left(\frac{1}{2} \csc^{-1}\left(\frac{x+b}{a}\right)\right) = \frac{x + b - \sqrt{(x + b)^2 - a^2}}{a}\tag{2}$$
These identities are generalizations of identities $(1-2)$ and $(9-10)$ at "Integration using Euler-like identities" and can be derived similarly as shown in the linked blog. The general identities $(1)$ and $(2)$ lead us to the following general transformation formula:
$$\boxed{\int f\left(x,\tan{\frac{\beta}{2}}, \tan{\frac{\gamma}{2}} \right)\,dx=\int f\left(\frac{e^{i\alpha}+e^{-i\alpha}}{2}a, e^{\pm\text{i}\alpha}, \frac{1-e^{\pm\text{i}\alpha}}{1+e^{\pm\text{i}\alpha}}\right)\,\frac{e^{-\text{i}\alpha}-e^{\text{i}\alpha}}{2i}a\,d\alpha}\tag{3}$$
Where $\alpha=\cos^{-1}\left(\frac{x+b}{a}\right)$, $\beta=\csc^{-1}\left(\frac{x+b}{a}\right)$ and $\gamma=\sec^{-1}\left(\frac{x+b}{a}\right).$ Use the upper sign for the alternating signs $\mp$ when $\frac{x + b}{a} \geq 1$ and the lower sign when $0 \leq \frac{x + b}{a} \leq 1$.
To illustrate its usefulness, let's look at the following example taken from the YouTube channel Prime Newtons.
Example 1. Evaluate
$$\int \frac{\sqrt{x+1}}{\sqrt{x+2}}\,dx$$
Solution. First, we obtain the values of $a$ and $b$ by solving the following (simple!) system of equations:
$$\left. b-a=1\atop a+b=2\right\}$$
The solutions are $a = \frac{1}{2}$ and $b = \frac{3}{2}$. Applying formula $(3)$ for $x\geq-1$, the integral becomes
$$\begin{aligned}\int \frac{\sqrt{x+1}}{\sqrt{x+2}}\,dx&= -\frac{i}{4}\int \frac{(1 - e^{i\alpha}) (e^{-i\alpha} - e^{i\alpha})}{(e^{i \alpha} + 1)} \, d\alpha\\&=-\frac{i}{4}\int \frac{e^{-i\alpha} (e^{i\alpha} - 1) (e^{2i\alpha} - 1)}{e^{i\alpha} + 1} \, dx\\&=-\frac{i}{4}\int e^{-i\alpha}(e^{i\alpha}-1)^2\,d\alpha\\&=-\frac{i}{4}\int (e^{-i\alpha}+e^{i\alpha}-2)\,d\alpha\\&=\frac14(2i\alpha+e^{-i\alpha}-e^{i\alpha})+C\end{aligned}$$
To switch to real numbers, replace from $(2)$, then replace the values of $a$ and $b$ and simplify, obtaining
$$=\sqrt{x^2+3x+2} + \frac{1}{2} \ln\left|2x - 2\sqrt{x^2+3x+2} + 3\right|+C$$
On the Prime Newtons YouTube channel, they first perform a variable change, followed by a trigonometric substitution, reducing the integral to integrals involving the secant and secant cubed. This integral calculator uses $u^2 = \frac{x+1}{x+2}$ and then (a complicated) partial fraction decomposition. This can be even more complicated for integral like this one:
$$\int \frac{x\sqrt{x+1}}{\sqrt{x+2}}\,dx$$
where as our method only requires a few lines of algebra and basic calculus, as in Example 1.
jueves, 21 de marzo de 2024
A new integration technique via Euler-like identities
"Complexification formulas are great and it seems like this simplifies the right away."- Ninad Munshi
We introduce Exponential Substitution, which is a transformation that simplifies certain composite trigonometric integrals and offers an alternative approach to integrating through trigonometric, hyperbolic, and Euler substitutions. Examples 2, 3, 7, 8, 9, 10 and 11 illustrate how this method could be more efficient than the traditional approaches.
Some Euler-like identities
The following identities have been suggested based on formulas in this blog post.
Theorem 1. If complex $\alpha=\cos^{-1}(x)$ and $\beta=\csc^{-1}(x)$, where real $x\in[1, \infty)$, then the following relation holds:
\[\boxed{e^{i\alpha}=\tan\left(\frac{\beta}{2}\right)}\tag{1}\]
In the complex plane, for real $x\in[0, 1]$, we have
\[\boxed{e^{-i\alpha}=\tan\left(\frac{\beta}{2}\right)}\tag{2}\]
Proof. Consider the left-hand side of equation $(1)$. By Euler's formula and properties of inverse trigonometric functions, the exponential $e^{i\cos^{-1}(x)}$ can be rewritten as follows.
\[e^{i\cos^{-1}{(x)}}=\cos{(\cos^{-1}{(x)})}+i\sin{(\cos^{-1}{(x)})}=x+i\sqrt{1-x^2}\]
The right-hand side of $(1)$ can be written as follows.
\[\tan{\left(\frac{\csc^{-1}{(x)}}{2}\right)}=\frac{\sin{(\csc^{-1}{(x)}})}{1+\cos{(\csc^{-1}{(x)}})}=\frac{\frac{1}{x}}{1+\frac{\sqrt{x^2-1}}{x}}=x-\sqrt{x^2-1}\]
Now we just need to find the solution set for non-negative real $x$ of the following complex equation.
\[x+i\sqrt{1-x^2}=x-\sqrt{x^2-1}\tag{3}\]
Clearly, equation $(3)$ is true for $x=1$. For $x>1$, it follows that $1-x^2<0$, and therefore, the left-hand side of $(3)$ simplifies to $x-\sqrt{x^2-1}$. Consequently, equation $(3)$ is true for $x\geq1$.
For identity $(2)$, equation $(3)$ becomes
\[x-i\sqrt{1-x^2}=x-\sqrt{x^2-1}\tag{4}\]
If $0 \leq x \leq 1$, then $x^2 - 1 < 0$. Therefore, the right-hand side of equation $(4)$ can be rewritten as $x - i\sqrt{1 - x^2}$, making equation $(4)$ valid for $[0, 1]$ in the complex plane.
$\square$
The functions on both sides of identity $(1)$ are continuous for $x \geq 1$. And for identity $(2)$, continuity is guaranteed on $[0, 1]$ for complex numbers.
Remark. If complex $\alpha=\cos^{-1}(x)$ and $\beta=\csc^{-1}(x)$, where real $x\in[-1, 1]$, then the following relation holds:
\[\boxed{e^{i\alpha}=\cot\left(\frac{\beta}{2}\right)}\tag{5}\]
This is true because identity $(5)$ can be rewritten as:
\[x+i\sqrt{1-x^2}=x+\sqrt{x^2-1}\tag{6}\]
And the solution set of $(6)$ is $x \in [-1, 1]$.
Theorem 2. If complex $\alpha=\cos^{-1}(x)$ and $\gamma=\sec^{-1}(x)$, where real $x\in [1, \infty)$, then the following relation holds:
\[\boxed{e^{i\alpha}=\frac{1-\tan\left(\frac{\gamma}{2}\right) }{1+\tan\left(\frac{\gamma}{2}\right)}}\tag{7}\]In the complex plane, for $x\in [0, 1]$, we have
\[\boxed{e^{-i\alpha}=\frac{1-\tan\left(\frac{\gamma}{2}\right) }{1+\tan\left(\frac{\gamma}{2}\right)}} \tag{8}\]
Proof. By property of inverse trigonometric functions we have
\[\tan{\left(\frac{\sec^{-1}(x)}{2}\right)}=\frac{\sin{(\sec^{-1}(x))}}{1+\cos{(\sec^{-1}(x))}}=\frac{\frac{\sqrt{x^2-1}}{x}}{1+\frac{1}{x}}=\frac{\sqrt{x-1}}{\sqrt{x+1}}\]Substituting into the right-hand side of identity $(7)$, we obtain
$$\frac{1-\tan\left(\frac{\gamma}{2}\right)}{1+\tan\left(\frac{\gamma}{2}\right)}=\frac{1-\frac{\sqrt{x-1}}{\sqrt{x+1}}}{1+\frac{\sqrt{x-1}}{\sqrt{x+1}}}=x-\sqrt{x^2-1}$$
Hence, the same argument for Theorem 1 applies to Theorem 2, and both identities $(7)$ and $(8)$ are continuous on their respective intervals, with identity $(8)$ being valid in the complex plane.
$\square$
For $x\in[1,\infty)$, identity $(7)$ can be rewritten as follows:
\[\boxed{\tan{\left(\frac{\gamma}{2}\right)}=\frac{1-e^{i\alpha}}{1+e^{i\alpha}}}\tag{9}\]
And in the complex plane, for $x\in[0, 1]$, identity $(8)$ can be rewritten as follows:
\[\boxed{\tan{\left(\frac{\gamma}{2}\right)}=\frac{1-e^{-i\alpha}}{1+e^{-i\alpha}}}\tag{10}\]
Remark. For similar reasons to those given for identity $(5)$, for \(x \in [-1, 1]\), the following identity holds:
\[\boxed{\cot{\left(\frac{\gamma}{2}\right)}=\frac{1+e^{i\alpha}}{1-e^{i\alpha}}}\tag{11}\]
Applications
Example 1. Evaluate \[\int_{1}^{2}\sqrt[3]{\tan{\left(\frac12\csc^{-1}(x)\right)}} \, dx\tag{12}\]
Solution. Using identity $(1)$, integral $(12)$ becomes:
\[\int_{1}^{2} e^{\frac{1}{3}i\alpha}dx\]
Since $\alpha=\cos^{-1}(x)\implies d\alpha=-\frac{1}{\sqrt{1-x^2}}\,dx$ and $x=\cos{(\alpha)}$.
$$\begin{aligned}\int_{1}^{2} e^{\frac{1}{3}i\alpha}dx&= -\int_{1}^{2} e^{\frac{1}{3}i\alpha}\sqrt{1-\cos^2{(\alpha)}}\,d\alpha\\&=-\int_{1}^{2} e^{\frac{1}{3}i\alpha}\sin{(\alpha)}\,d\alpha\\&=-\frac{i}{2}\int_{1}^{2}(e^{-\frac23i\alpha}- e^{\frac43i\alpha})\,d\alpha\qquad (\text{Since $\sin{(\alpha)}=\frac12i(e^{-i\alpha}-e^{i\alpha}$}).)\\&=\frac34e^{-\frac23i\alpha}+\frac38e^{\frac43i\alpha}\bigg|_1^2\\&=\frac38e^{-\frac23i\alpha}(e^{2i\alpha}+2)\bigg|_1^2\\&=\frac{3\left((x-\sqrt{x^2-1})^2+2\right)}{8(x-\sqrt{x^2-1})^{2/3}}\bigg|_1^2\qquad (\text{Changing back to reals.})\\&=\frac{{3(2 + (2 - \sqrt{3})^2)}}{{8(2 - \sqrt{3})^{\frac{2}{3}}}} - \frac{9}{8}\\&\approx0.744\end{aligned}$$
Another approach involves using the simplification $\tan{\left(\frac12\csc^{-1}{(x)}\right)}=x-\sqrt{x^2-1}$. However, this method could lead to more complicated integrals, especially when evaluating rational trigonometric integrals with multiple terms involving $\tan{\left(\frac12\csc^{-1}{(x)}\right)}$ or $\tan{\left(\frac12\sec^{-1}{(x)}\right)}$ or both. Making an appropriate u-substitution might not be immediately obvious in those cases, whereas using identities $(1)$ and $(9)$ quickly turns the integral into an integral of a rational function. As an illustration, let's consider the following example.
Example 2. Evaluate
\[\int\frac{1}{\tan\left(\frac{1}{2}\csc^{-1} (x)\right) - \tan\left(\frac{1}{2}\sec^{-1}(x)\right)} \, dx\]
Solution. First, we rewrite the integral using the identity $(1)$ and $(9)$:
\[\int \frac{1}{e^{i\alpha} - \frac{1 - e^{i\alpha}}{1 + e^{i\alpha}}} \, dx=\int\frac{1+e^{i\alpha}}{e^{2i\alpha}+2e^{i\alpha}-1}\,dx\]Similar to how we proceeded in example 1, since $\alpha=\cos^{-1}(x)\implies d\alpha=-\frac{1}{\sqrt{1-x^2}}\,dx$ and $x=\cos{(\alpha)}$. Then
$$\begin{aligned}\int\frac{1+e^{i\alpha}}{e^{2i\alpha}+2e^{i\alpha}-1}\,dx &=-\int\frac{1+e^{i\alpha}}{e^{2i\alpha}+2e^{i\alpha}-1}\sqrt{1-\cos^2(\alpha)}\,d\alpha\\&=-\int\frac{1+e^{i\alpha}}{e^{2i\alpha}+2e^{i\alpha}-1}\sin(\alpha)\,d\alpha\\&=-\frac{i}{2}\int\frac{e^{-i\alpha} - e^{i\alpha} - e^{2i\alpha} + 1}{e^{2i\alpha}+2e^{i\alpha}-1}\,d\alpha\end{aligned}$$
Let $u=e^{i\alpha}\implies du=ie^{i\alpha}\,d\alpha$. Then, putting terms over a common denominator, and factoring, we have
$$-\frac{i}{2}\int\frac{e^{-i\alpha} - e^{i\alpha} - e^{2i\alpha} + 1}{e^{2i\alpha}+2e^{i\alpha}-1}\,d\alpha=\frac12\int\frac{(u - 1) (u + 1)^2}{u^2 (u^2 + 2 u - 1)}\,du$$
At this point, we can proceed by applying partial fraction decomposition.
Remark. In the MathSE forum, I asked about alternative approaches to the integral in example 2. Zacky has suggested the following substitution
$$\int \frac{1}{x-\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}}\,dx \overset{\frac{x-1}{x+1}=t^2}{=}4\int \frac{t}{(1+t)(1-t)^2(1-2t-t^2)}\,dt$$
However, as the reader may have noticed by the degree of the denominator, I would argue that this substitution leads us to a more complicated partial fraction decomposition. Look here and here and compare for yourself.
The following integral was asked by user SAQ on the MathSE forum.
Example 3. Evaluate
\[ \int\dfrac{\sqrt{x-1}-\sqrt{x+1}}{\sqrt{x-1}-3\sqrt{x+1}}\,dx\]
Solution. We first divide both the numerator and denominator by $\sqrt{x+1}$:
\[\int\dfrac{\sqrt{x-1}-\sqrt{x+1}}{\sqrt{x-1}-3\sqrt{x+1}}\,dx = \int\dfrac{\frac{\sqrt{x-1}}{\sqrt{x+1}}-1}{\frac{\sqrt{x-1}}{\sqrt{x+1}}-3}\,dx\]Now, we'll substitute $\frac{\sqrt{x-1}}{\sqrt{x+1}}$ with $\frac{1-e^{i\alpha}}{1+e^{i\alpha}}$ using the identity $(9)$:
\[= \int\dfrac{\frac{1-e^{i\alpha}}{1+e^{i\alpha}}-1}{\frac{1-e^{i\alpha}}{1+e^{i\alpha}}-3}\,dx\]As $\alpha = \cos^{-1}(x)$, then $x = \cos{\alpha}$ and $dx = -\sqrt{1-x^2}\,d\alpha = -\sqrt{1-\cos^2{(\alpha)}}\, d\alpha = -\sin{(\alpha)}\,d\alpha$. Now, we'll perform the substitution:
\[= -\int\dfrac{\frac{1-e^{i\alpha}}{1+e^{i\alpha}}-1}{\frac{1-e^{i\alpha}}{1+e^{i\alpha}}-3}\sin{(\alpha)}\,d\alpha\]Since $\sin(\alpha) = \frac{i}{2}(e^{-i\alpha} - e^{i\alpha})$ and simplifying, we get:
\[\frac{i}{2}\int \frac{e^{2i\alpha}-1}{2e^{i\alpha}+1} \, d\alpha\]Now, let $u = e^{i\alpha}$. Then, $du = ie^{i\alpha}\,d\alpha = iu\,d\alpha$ and $d\alpha = \frac{1}{iu}\,du$. Substituting, performing long division, and then decomposing into partial fractions,
\[ \begin{aligned}\frac12\int \frac{u^2 - 1}{u(2u + 1)} \, du&=\frac14\int 1\,du - \frac14\int \frac{u+2}{u(2u+1)}\,du\\&=\frac14\int1\,du -\frac12\int\frac{1}{u}\,du+\frac38\int\frac{1}{2u+1}\,du\\&=\frac{u}{4}-\frac{\ln{u}}{2}+\frac38\ln{|2u+1|}+C\\&=\frac{u}{4}+\frac12\ln{\left|\frac{(2u+1)^{\frac34}}{u}\right|}+C\\&=\frac{e^{i\alpha}}{4}+\frac12\ln{\left|\frac{(2e^{i\alpha}+1)^{\frac34}}{e^{i\alpha}}\right|}+C\\&=\frac{x-\sqrt{x^2-1}}{4}+\frac12\ln{\left|\frac{(2(x-\sqrt{x^2-1})+1)^{\frac34}}{x-\sqrt{x^2-1}}\right|}+C\end{aligned} \]
The following example has been taken from the popular YouTube channel, Blackpenredpen.
Example 4. Evaluate
$$\int\sqrt{\frac{1-x}{1+x}}\,dx$$
Solution. For $x\in[0, 1]$, applying identity $(10)$, we have
$$\begin{aligned}\int\sqrt{\frac{1-x}{1+x}}\,dx&=i\int\sqrt{\frac{x-1}{x+1}}\,dx\\&=-i\int\frac{1-e^{-i\alpha}}{1+e^{-i\alpha}}\sin(\alpha)\,d\alpha\\&=\frac12\int\frac{(e^{-i\alpha}-1)(e^{-i\alpha}-e^{i\alpha})}{1+e^{-i\alpha}}\,d\alpha\\&=\frac12\int((e^{-i\alpha}+e^{i\alpha})-2)\,d\alpha\\&=\frac12\int2(\cos(\alpha)-1)\,d\alpha\\&=\sin(\alpha)-\alpha+C\\&=\sqrt{1-x^2}-\cos^{-1}(x)+C\end{aligned}$$Other integration techniques
miércoles, 28 de febrero de 2024
A family of trigonometric formulas for the roots of quadratic equations
This note presents alternative trigonometric formulas for finding the roots of quadratic equations where $a$, $b$, and $c$ are non-zero real numbers.
Before the era of calculators, trigonometric formulas were favored for computing quadratic roots due to their time and labor-saving benefits. However, it's not advisable nowadays to rely on these formulas over the traditional quadratic formula for root calculations. Nonetheless, the trigonometric formulas showcased in this note have hinted at certain identities that appear useful for integrating trigonometric functions. Thus, the sole purpose of this note is to document the origins of these identities.
Theorem 1. Let $a$, $b$, and $c$ be non-zero real numbers. For the quadratic equation $ax^2+bx+c=0$, the roots are given by:
$$x_{1}= ie^{i\alpha}\sqrt{\frac{c}{a}}\qquad \text{and}\qquad x_2=-ie^{-i\alpha}\sqrt{\frac{c}{a}},\tag{1}$$
where $\alpha=\sin^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.
Proof. Consider the quadratic equation $a^2+bx+c=0$. Multiplying both sides by $a$ and making the substitutions $ax=p$ and $ac=q^2$ yields
$$p^2+bp+q^2=0.$$
Let's make the substitution $\sin{\alpha} = \frac{b}{2q}$. Consequently,
$$\begin{aligned}0&=p^2+bp+q^2\\&=p^2+2qp\sin{\alpha}+q^2\\&=p^2\left(\sin^2{\frac{\alpha}{2}}+\cos^2{\frac{\alpha}{2}}\right)+4qp\sin{\frac{\alpha}{2}}\cos{\frac{\alpha}{2}}+q^2\left(\sin^2{\frac{\alpha}{2}}+\cos^2{\frac{\alpha}{2}}\right)\\&=\left(p\sin{\frac{\alpha}{2}}+q\cos{\frac{\alpha}{2}}\right)^2 + \left(p\cos{\frac{\alpha}{2}}+q\sin{\frac{\alpha}{2}}\right)^2\\&=\left(p\sin{\frac{\alpha}{2}}+q\cos{\frac{\alpha}{2}}\right)^2 -i^2\left(p\cos{\frac{\alpha}{2}}+q\sin{\frac{\alpha}{2}}\right)^2 .\end{aligned}$$
Factorizing the difference of squares and then factorizing again, one of the factor of the quadratic equation can be express as follows
$$p\left(\sin{\frac{\alpha}{2}}+i\cos{\frac{\alpha}{2}}\right)+q\left(\cos{\frac{\alpha}{2}}+i\sin{\frac{\alpha}{2}}\right).$$
Setting the factor equal to zero, undoing the substitutions $ax=p$ and $ac=q^2$ and solving for $x$, we obtain
$$\begin{aligned}x_1&=-\left(\frac{\cos{\frac{\alpha}{2}}+i\sin{\frac{\alpha}{2}}}{\sin{\frac{\alpha}{2}}+i\cos{\frac{\alpha}{2}}}\right)\sqrt{\frac{c}{a}}\\&= -(\sin{\alpha}-i\cos{\alpha})\sqrt{\frac{c}{a}}\\&=ie^{i\alpha}\sqrt{\frac{c}{a}}. \end{aligned}$$
The other factor is given by
$$p\left(\sin{\frac{\alpha}{2}}-i\cos{\frac{\alpha}{2}}\right)+q\left(\cos{\frac{\alpha}{2}}-i\sin{\frac{\alpha}{2}}\right).$$
And similarly,
$$x_2=-ie^{-i\alpha}\sqrt{\frac{c}{a}}.$$
Theorem 2. For the quadratic equation $ax^2+bx+c=0$ with non-zero real numbers $a$, $b$, and $c$, the roots are given by:
$$x_{1}=-\tan{\frac{\beta}{2}}\sqrt{\frac{c}{a}}\qquad \text{and} \qquad x_{2}=-\cot{\frac{\beta}{2}}\sqrt{\frac{c}{a}},\tag{2}$$
where $\beta=\csc^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.
Proof. Multiplying by $a$ the quadratic equation $ax^2+bx+c=0$ and then substituting $ax=p$ and $ac=q^2$, we have
$$p^2+bp+q^2=0.$$
Use the substitution $\csc{\beta}=\frac{b}{2q}$, then
$$\begin{aligned}0&=p^2+bp+q^2\\&=p^2+2qp\csc{\beta}+q^2\\&=\csc{\beta}\left(p^2\sin{\beta}+2qp+q^2\sin{\beta}\right)\\&=2p^2\sin{\frac{\beta}{2}}\cos{\frac{\beta}{2}}+2qp\left(\sin^2{\frac{\beta}{2}}+\cos^2{\frac{\beta}{2}}\right)+2q^2\sin{\frac{\beta}{2}}\cos{\frac{\beta}{2}}\\&=p^2\sin{\frac{\beta}{2}}\cos{\frac{\beta}{2}}+qp\sin^2{\frac{\beta}{2}}+ qp\cos^2{\frac{\beta}{2}}+q^2\sin{\frac{\beta}{2}}\cos{\frac{\beta}{2}}\\&= p\sin{\frac{\beta}{2}}\left(p\cos{\frac{\beta}{2}}+q\sin{\frac{\beta}{2}}\right)+q\cos{\frac{\beta}{2}}\left(p\cos{\frac{\beta}{2}}+q\sin{\frac{\beta}{2}}\right)\\&= \left(p\cos{\frac{\beta}{2}}+q\sin{\frac{\beta}{2}}\right)\left(p\sin{\frac{\beta}{2}}+q\cos{\frac{\beta}{2}}\right).\end{aligned}$$By setting the factors equal to zero, undoing substitutions $ax=p$ and $ac=q^2$ and solving for $x$, we obtain the desired formulas in $(2)$.
Theorem 3. Let $a$, $b$ and $c$ be non-zero real numbers. If $ax^2+bx+c=0$, then the roots are given by:
$$x_{1}= -e^{i\gamma}\sqrt{\frac{c}{a}}\qquad \text{and}\qquad x_2=-e^{-i\gamma}\sqrt{\frac{c}{a}},\tag{3}$$
where $\gamma=\cos^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.
Proof. Consider the quadratic equation $ax^2+bx+c=0$. Multiplying both sides by $a$ and substituting $ax=p$ and $ac=q^2$ yields
$$p^2+bp+q^2=0.$$
By making the substitution $\cos{\gamma} = \frac{b}{2q}$, it follows that
$$\begin{aligned}0&=p^2+bp+q^2\\&=p^2+2qp\cos{\gamma}+q^2\\&=p^2\left(\sin^2{\frac{\gamma}{2}}+\cos^2{\frac{\gamma}{2}}\right)+4qp\left(\cos^2{\frac{\gamma}{2}}-\sin^2{\frac{\gamma}{2}}\right)+q^2\left(\sin^2{\frac{\gamma}{2}}+\cos^2{\frac{\gamma}{2}}\right)\\&=\sin^2{\frac{\gamma}{2}} (p-q)^2+\cos^2{\frac{\gamma}{2}}(p+q)^2\\&=\sin^2{\frac{\gamma}{2}} (p-q)^2-i^2\cos^2{\frac{\gamma}{2}}(p+q)^2.\end{aligned}$$
By factoring the difference of squares first and then applying further factorization, one of the factors of the quadratic equation can be represented as follows
$$p\left(\sin{\frac{\gamma}{2}}+i\cos{\frac{\gamma}{2}}\right)+q\left(-\sin{\frac{\gamma}{2}}+i\cos{\frac{\gamma}{2}}\right).$$
Setting the factors equal to zero, undoing substitutions $ax=p$ and $ac=q^2$ and solving for $x$, we obtain
$$\begin{aligned}x_1&=\left(\frac{\sin{\frac{\gamma}{2}}-i\cos{\frac{\gamma}{2}}}{\sin{\frac{\gamma}{2}}+i\cos{\frac{\gamma}{2}}}\right)\sqrt{\frac{c}{a}}\\&= -(\cos{\gamma}+i\sin{\gamma})\sqrt{\frac{c}{a}}\\&=-e^{i\gamma}\sqrt{\frac{c}{a}}.\end{aligned}$$
The other factor is given by
$$p\left(\sin{\frac{\gamma}{2}}-i\cos{\frac{\gamma}{2}}\right)-q\left(\sin{\frac{\gamma}{2}}+i\cos{\frac{\gamma}{2}}\right).$$
And similarly,
$$x_2=-e^{-i\gamma}\sqrt{\frac{c}{a}}.$$
Theorem 4. Let $a$, $b$ and $c$ be non-zero real numbers. If $ax^2+bx+c=0$, then the roots are given by:
$$x_{1,2}=\frac{\tan{\frac{\delta}{2}\pm1}}{\tan{\frac{\delta}{2}\mp1}}\sqrt{\frac{c}{a}},\tag{4}$$
where $\delta=\sec^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.
Proof. Consider the quadratic equation $ax^2+bx+c=0$. Multiplying both sides by $a$ and substituting $ax=p$ and $ac=q^2$ yields
$$p^2+bp+q^2=0.$$
Use the substitution $\sec{\delta}=\frac{b}{2q}$, then
$$\begin{aligned}0&=p^2+bp+q^2\\&=p^2+2qp\sec{\delta}+q^2\\&=\sec{\delta}\left(p^2\cos{\delta}+2qp+q^2\cos{\delta}\right)\\&=\cos{\delta}(p^2+q^2)+2qp\\&= \cos{\delta}(p+q)^2-2qp\cos{\delta}+2qp\\&= \left(\cos^2{\frac{\delta}{2}}-\sin^2{\frac{\delta}{2}}\right)(p+q)^2-2qp\left(1-2\sin^2{\frac{\delta}{2}}\right)+2qp\\&= \cos^2{\frac{\delta}{2}}(p+q)^2 -\sin^2{\frac{\delta}{2}}(p+q)^2 +4qp\sin^2{\frac{\delta}{2}}\\&= \cos^2{\frac{\delta}{2}}(p+q)^2 -\sin^2{\frac{\delta}{2}}\left((p+q)^2 -4qp\right)\\&= \cos^2{\frac{\delta}{2}}(p+q)^2 -\sin^2{\frac{\delta}{2}}(p-q)^2\\&= \left(\cos{\frac{\delta}{2}}(p+q) +\sin{\frac{\delta}{2}}(p-q)\right) \left(\cos{\frac{\delta}{2}}(p+q) -\sin{\frac{\delta}{2}}(p-q)\right).\end{aligned}$$
Expanding and then factorizing again, one of the factors of the quadratic equation is given by
$$p\left(\cos{\frac{\delta}{2}}+\sin{\frac{\delta}{2}}\right)+q\left(\cos{\frac{\delta}{2}}-\sin{\frac{\delta}{2}}\right).$$
By setting the factors equal to zero, undoing substitutions $ax=p$ and $ac=q^2$ and solving for $x$, we obtain
$$\begin{aligned}x_1&=\frac{\sin{\frac{\delta}{2}}+\cos{\frac{\delta}{2}}}{\sin{\frac{\delta}{2}}-\cos{\frac{\delta}{2}}}\sqrt{\frac{c}{a}}\\&= \frac{\tan{\frac{\delta}{2}+1}}{\tan{\frac{\delta}{2}-1}}\sqrt{\frac{c}{a}}. \end{aligned}$$
The other factor is given by
$$p\left(\cos{\frac{\delta}{2}}-\sin{\frac{\delta}{2}}\right)+q\left(\cos{\frac{\delta}{2}}+\sin{\frac{\delta}{2}}\right).$$
Similarly,
$$x_2=\frac{\tan{\frac{\delta}{2}-1}}{\tan{\frac{\delta}{2}+1}}\sqrt{\frac{c}{a}}.$$
Theorem 5. Let $a$, $b$ and $c$ be non-zero real numbers. If $ax^2+bx+c=0$, then the roots are given by:
$$x_{1,2}=\frac{i\pm e^{\eta}}{i\mp e^{\eta}}\sqrt{\frac{c}{a}},\tag{5}$$
where $\eta=\tanh^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.
Proof. Consider the quadratic equation $ax^2+bx+c=0$. Multiplying both sides by $a$ and substituting $ax=p$ and $ac=q^2$ yields
$$p^2+bp+q^2=0.$$
Use the substitution $\tanh{\eta}=\frac{b}{2q}$, then
$$\begin{aligned}0&=p^2+bp+q^2\\&=p^2+2qp\tanh{\eta}+q^2\\&=(e^{2\eta} + 1) p^2 + 2 (e^{2\eta} - 1) p q + (e^{2\eta} + 1) q^2\\&=e^{2\eta}(p + q)^2 + (p - q)^2\\&=e^{2\eta}(p + q)^2 - i^2(p - q)^2 \\&=(e^{\eta}p + i p + e^{\eta}q - i q)(e^{\eta}p - i p + e^{\eta}q + i q)\\&= \left(p(e^{\eta} + i) + q(e^{\eta} - i)\right)\left(p(e^{\eta} - i) + q(e^{\eta} + i)\right)\end{aligned}$$
Setting the factor equal to zero, undoing the substitutions $ax=p$ and $ac=q^2$ and solving for $x$, we obtain
$$x_{1}=\frac{i- e^{\eta}}{i+e^{\eta}}\sqrt{\frac{c}{a}}.$$
And similarly,
$$x_{2}=\frac{i+ e^{\eta}}{i-e^{\eta}}\sqrt{\frac{c}{a}}.$$
Theorem 6. Let $a$, $b$ and $c$ be non-zero real numbers. If $ax^2+bx+c=0$, then the roots are given by:
$$x_{1,2}=\frac{1\pm e^{\theta}}{1\mp e^{\theta}}\sqrt{\frac{c}{a}},\tag{6}$$
where $\theta=\coth^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.
Proof. Consider the quadratic equation $ax^2+bx+c=0$. Multiplying both sides by $a$ and substituting $ax=p$ and $ac=q^2$ yields
$$p^2+bp+q^2=0.$$
Use the substitution $\coth{\theta}=\frac{b}{2q}$, Then, similar to how we proceeded previously,
$$\begin{aligned}0&=p^2+bp+q^2\\&=p^2+2qp\coth{\theta}+q^2\\&=(e^{2\theta} - 1) p^2 + 2 (e^{2\theta} + 1) p q + (e^{2\theta} - 1) q^2\\&=e^{2\theta}(p + q)^2 - (p - q)^2 \\&=\left( p e^{\theta} + p + q e^{\theta} - q \right) \left( p e^{\theta} - p + q e^{\theta} + q \right)\\&=\left(p(e^x + 1) + q(e^x - 1)\right)\left(p (e^x - 1) + q (e^x + 1)\right)\end{aligned}$$
Setting the factor equal to zero, undoing the substitutions $ax=p$ and $ac=q^2$ and solving for $x$, we obtain
$$x_{1}=\frac{1- e^{\theta}}{1+e^{\theta}}\sqrt{\frac{c}{a}}= -\tanh\left(\frac{1}{2} \text{coth}^{-1}(x)\right)\sqrt{\frac{c}{a}}.$$
And similarly,
$$x_{2}=\frac{1+ e^{\theta}}{1-e^{\theta}}\sqrt{\frac{c}{a}}.$$
jueves, 15 de febrero de 2024
Integrals yielding $e^{\pi}$ or $e^{-\pi}$
Lately, I've been playing a lot with integrals, and coincidentally (with a bit of algebraic manipulation), I've come across these two beauties:
$$\int_{0}^{1} \left(\frac{5}{2} \left((x - \sqrt{x^2 - 1})^{2i} + x^4\right) - 1\right) \, dx = e^{\pi},\tag{1}$$
$$\int_{0}^{1} \left( -\frac{5}{2\left( x - \sqrt{x^2 - 1}\right)^{2i}} - \frac{5x^4}{2} + 1 \right) \, dx = e^{-\pi}.\tag{2}$$
$e^{\pi}$ is known as Gelfond's constant.
I have provided these integrals as a response to a question on MathSE. The proofs are left as exercises for the reader.
miércoles, 7 de febrero de 2024
Solving 'impossible' integrals with a new trick
"Complexification formulas are great and it seems like this simplifies the right away."Ninad Munshi
If complex $\theta_1=\cos^{-1}(p)$ and $\theta_2=\sec^{-1}(p)$, where $p\in(-1, 0) \cup (1, \infty)$, then the following relation holds:
$$e^{i\theta_1}=\frac{1-\tan\frac{\theta_2}{2} }{1+\tan\frac{\theta_2}{2}}.
\tag{1}\label{463459_1}$$
And for $p\in(-\infty, -1)\cup (0, 1)$, we have
$$e^{-i\theta_1}=\frac{1-\tan\frac{\theta_2}{2} }{1+\tan\frac{\theta_2}{2}}. \tag{2}\label{463459_2}$$
If complex $\theta_1=\sin^{-1}(p)$ and $\theta_2=\csc^{-1}(p)$, where $p\in(-\infty, -1)\cup (0, 1)$, then the following relation holds:
$$ie^{i\theta_1}=-\tan\frac{\theta_2}{2}.\tag{3}\label{463459_3}$$
And for $p\in(-1, 0) \cup (1, \infty)$ we have
$$ie^{-i\theta_1}=\tan\frac{\theta_2}{2}.\tag{4}\label{463459_4}$$
There are several variants that we can obtain by equating (and simplifying) the trigonometric formulas for quadratic equations from my previous question.
I have noticed that for certain trigonometric integrals defined over permissible intervals of $p$, the evaluation simplifies considerably. For instance, consider the following definite integral:
$$\int_2^5 \sqrt{\tan\left(\frac{\csc^{-1}(x)}{2}\right)} \,dx.\tag{5}$$
This integral calculator returns the following (although Wolfram Alpha solves it):
No antiderivative could be found within the given time limit, or all supported integration methods were tried unsuccessfully. Note that many functions don't have an elementary antiderivative.
But it gives an approximation of $1.178881841955109.$
Given that the interval $[2, 5]$ is within the permissible values of $p$, we can use $e^{i\cos^{-1}(p)}=\tan\frac{\csc^{-1}(p)}{2}$ (derived from identities $(1)$ and $(4)$, valid for $p\in[-1, 0) \cup [1, \infty)$, to convert $(5)$ into
$$\int_2^5 \sqrt{e^{i\cos^{-1}(x)}}\,dx.\tag{6}$$
The same calculator provides the same answer but now displaying the steps as well. As a second example, Mathematica is unable to solve this integral: