Ahmed’s Integral

March 27, 2020 / 1 Comment on Ahmed’s Integral

Prove that

    \[\int_{0}^{1} \frac{\arctan\sqrt{x^{2} + 2}}{\sqrt{x^{2} + 2}} \, \frac{\mathrm{d}x}{x^{2}+1} = \frac{5\pi^{2}}{96}\]

Solution

Consider the function \displaystyle{f(t) = \int_{0}^{1} \frac{\arctan \left(t\sqrt{2+x^2} \right)}{(1+x^2)\sqrt{2+x^2}} \, \mathrm{d}x}. Differentiating with respect to t we have that:

    \begin{align*} f'(t) &= \frac{\mathrm{d} }{\mathrm{d} t} \int_{0}^{1} \frac{\arctan \left(t\sqrt{2+x^2} \right)}{(1+x^2)\sqrt{2+x^2}} \, \mathrm{d}x \\ &= \int_{0}^{1} \frac{\partial }{\partial t} \frac{\arctan \left(t\sqrt{2+x^2} \right)}{(1+x^2)\sqrt{2+x^2}} \, \mathrm{d}x \\ &= \int_{0}^{1} \frac{\mathrm{d}x}{\left ( 1+x^2 \right ) \left ( 1+2t^2 + t^2 x^2 \right )}\\ &= \int_{0}^{1} \frac{\mathrm{d}x}{\left ( 1+t^2 \right )\left ( 1+x^2 \right )} - \int_{0}^{1} \frac{t^2}{\left ( 1+t^2 \right ) \left ( 1+2t^2+t^2 x^2 \right )} \, \mathrm{d}x \\ &= \frac{\pi}{4} \cdot \frac{1}{1+t^2} - \frac{t}{\left ( 1+t^2 \right ) \sqrt{1+2t^2}} \arctan \frac{t}{\sqrt{1+2t^2}} \end{align*}

We integrate the last equation from 1 to \infty. Thus,

    \begin{align*} \int_{1}^{\infty} f'(t) \, \mathrm{d}t &= \int_1^\infty \left (\frac{\pi}{4} \cdot \frac{1}{1+t^2} - \frac{t}{\left ( 1+t^2 \right ) \sqrt{1+2t^2}} \arctan \frac{t}{\sqrt{1+2t^2}} \right ) \, \mathrm{d}t \\ &=\frac{\pi}{4} \left ( \frac{\pi}{2} - \frac{\pi}{4} \right ) - \int_1^\infty \frac{t}{\left ( 1+t^2 \right ) \sqrt{1+2t^2}} \arctan \frac{t}{\sqrt{1+2t^2}} \, \mathrm{d}t \\ &= \frac{\pi^2}{16} - \int_1^\infty \frac{t}{\left ( 1+t^2 \right ) \sqrt{1+2t^2}} \arctan \frac{t}{\sqrt{1+2t^2}} \, \mathrm{d}t \end{align*}

However,

    \begin{align*} \lim_{t \rightarrow +\infty} f(t) &= \lim_{t \rightarrow +\infty} \int_{0}^{1} \frac{\arctan \left(t\sqrt{2+x^2} \right)}{(1+x^2)\sqrt{2+x^2}} \, \mathrm{d}x \\ &= \int_{0}^{1} \lim_{t \rightarrow +\infty} \frac{\arctan \left(t\sqrt{2+x^2} \right)}{(1+x^2)\sqrt{2+x^2}} \, \mathrm{d}x \\ &= \frac{\pi}{2} \int_{0}^{1} \frac{\mathrm{d}x}{\left ( 1+x^2 \right )\sqrt{2+x^2}}\\ &\!\!\!\!\!\!\!\overset{\text{Lemma 2}}{=\! =\! =\! =\! =\! =\! } \frac{\pi}{2} \cdot \frac{\pi}{6} \\ &= \frac{\pi^2}{12} \end{align*}

Hence the last equation gives

(1)   \begin{equation*} \frac{\pi^2}{12} - f(1) = \frac{\pi^2}{16} - \int_1^\infty \frac{t}{\left ( 1+t^2 \right ) \sqrt{1+2t^2}} \arctan \frac{t}{\sqrt{1+2t^2}} \, \mathrm{d}t \end{equation*}

Suffice to calculate the integral. Applying the change of variables t \mapsto \frac{1}{t} we have:

    \begin{align*} \mathcal{J} &= \int_1^\infty \frac{t}{\left ( 1+t^2 \right ) \sqrt{1+2t^2}} \arctan \frac{t}{\sqrt{1+2t^2}} \, \mathrm{d}t \\ &= \int_{1}^{\infty} \frac{1}{1+\left ( \frac{1}{t} \right )^2 \sqrt{2+\frac{1}{t^2}}} \arctan \frac{1}{\sqrt{2 + \frac{1}{t^2}}} \frac{\mathrm{d}t}{t^2} \\ &\!\!\!\!\overset{t \mapsto 1/t}{=\! =\! =\! =\!} \int_{0}^{1} \frac{1}{\left (1+t^2 \right )\sqrt{2+t^2}} \arctan \frac{1}{\sqrt{2+t^2}} \, \mathrm{d}t\\ &= \int_{0}^{1}\frac{1}{\left (1+t^2 \right )\sqrt{2+t^2}} \left ( \frac{\pi}{2} - \arctan \sqrt{2+t^2}\right ) \, \mathrm{d}t \\ &=\frac{\pi}{2} \int_{0}^{1} \frac{\mathrm{d}t}{\left (1+t^2 \right )\sqrt{2+t^2}} - f(1) \\ &= \frac{\pi}{2} \cdot \frac{\pi}{6} - f(1) \\ &= \frac{\pi^2}{12} - f(1) \end{align*}

Going back at (1) we have that:

    \begin{align*} \frac{\pi^2}{12} - f(1) = \frac{\pi^2}{16} - \left ( \frac{\pi^2}{12} - f(1) \right ) &\Leftrightarrow \frac{\pi^2}{12} - f(1) = \frac{\pi^2}{16} - \frac{\pi^2}{12} + f(1) \\ &\Leftrightarrow 2f(1) = \frac{5\pi^2}{48} \\ &\Leftrightarrow \boxed{f(1) = \frac{5 \pi^2}{96}} \end{align*}

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Coxeter’s Integral

March 26, 2020 / 1 Comment on Coxeter’s Integral

Prove that

    \[\int_{0}^{\pi/4}\arctan \sqrt{\frac{\cos 2\theta }{2\cos^2 \theta }} \, \mathrm{d}\theta = \frac{\pi^2}{24}\]

Solution

We state 3 lemmata:

Lemma 1: It holds that \displaystyle \arctan x = \int_{0}^{1} \frac{x}{1+x^2 t^2} \, \mathrm{d}t.

Lemma 2: It holds that \displaystyle \int_{0}^{1}\frac{\mathrm{d}x}{\left(x^2+1 \right)\sqrt{x^2+2}}=\frac{\pi }{6}.

Proof: We have successively:

    \begin{align*} \int_{0}^{1}\frac{\mathrm{d}x}{\sqrt{x^2+2}(x^2+1)}&\overset{x=\sqrt{2} \sinh t}{=\! =\! =\! =\! =\! =\! =\!} \int_{0}^{\operatorname{arcsinh} \frac{1}{\sqrt{2}}}\frac{\mathrm{d}t}{1+2\sinh^2 t} \\ &=\int_{0}^{\operatorname{arcsinh} \frac{1}{\sqrt{2}}} \frac{\mathrm{d}t}{\cosh 2t} \\ &=\int_{0}^{\operatorname{arcsinh} \frac{1}{\sqrt{2}}}\frac{\cosh 2t}{1+\sinh^2 2t} \, \mathrm{d}t \\ &=\frac{1}{2} \operatorname{arctanh} \left ( \sinh 2\left ( \operatorname{arcsinh} \frac{1}{2} \right ) \right) \\ &= \frac{1}{2} \operatorname{arctanh} \sqrt{3} \\ &= \frac{\pi}{6} \end{align*}

Lemma 3: It holds that \displaystyle \int_{0}^{\infty}\frac{dx}{(x^2+a^2)(x^2+b^2)}=\frac{\pi }{2ab(a+b)} where a , b \neq 0.

Proof: We have successively:

    \begin{align*} \int_{0}^{\infty }\frac{\mathrm{d}x}{(x^2+a^2)(x^2+b^2)} &=\int_{0}^{\infty }\frac{1}{b^2-a^2}\left ( \frac{1}{x^2+a^2}-\frac{1}{x^2+b^2} \right ) \, \mathrm{d} x \\ &=\frac{1}{b^2-a^2}\left ( \frac{\pi }{2a}-\frac{\pi }{2b} \right )\\ &=\frac{\pi }{2ab(a+b)} \end{align*}

We are ready to attack the initial monster. For that we have:

    \begin{align*} \int_{0}^{\pi /4}\arctan \sqrt{\frac{\cos 2\theta }{2\cos ^2\theta }} \, \mathrm{d}\theta &= \int_{0}^{\pi /4}\int_{0}^{1}\frac{\sqrt{\frac{\cos 2\theta }{2\cos^2 \theta }}}{1+\left(\frac{\cos 2\theta }{2\cos ^2\theta } \right)x^2} \, \mathrm{d}x \, \mathrm{d}\theta \\ &\!\!\!\!\!\!=\int_{0}^{1}\int_{0}^{\pi /4}\frac{\sqrt{2}\cos \theta\sqrt{1-2\sin ^2\theta }}{2-2\sin ^2\theta +(1-2\sin ^2 \theta )x^2} \, \mathrm{d}\theta \, \mathrm{d}x\\ &\!\!\!\!\!\!=\int_{0}^{1}\int_{0}^{\pi /4}\frac{\cos \varphi\sqrt{1-\sin ^2 \varphi }}{2-\sin ^2 \varphi+\left(1-\sin ^2\varphi )x^2 \right) } \, \mathrm{d}\varphi \, \mathrm{d}x \\ &=\int_{0}^{1}\int_{0}^{\pi /4}\frac{\cos ^2\varphi }{\sin ^2 \varphi +(x^2+2)\cos ^2\varphi }\, \mathrm{d}\varphi \, \mathrm{d}x\\ &=\int_{0}^{1}\int_{0}^{\pi /2}\frac{\mathrm{d}\varphi \, \mathrm{d}x}{\tan^2\varphi +x^2+2}\\ &\!\!\!\!\!\!\overset{y=\tan \varphi}{=\! =\! =\! =\! =\! =\!} \int_{0}^{1}\int_{0}^{\infty}\frac{\mathrm{d}y \, \mathrm{d}x}{(y^2+x^2+2)(y^2+1)}\\ &=\frac{\pi }{2}\int_{0}^{1}\frac{\mathrm{d}y}{(1+\sqrt{2+y^2})\sqrt{2+y^2}}\\ &=\frac{\pi }{2}\left(\frac{\pi }{4}-\frac{\pi }{6} \right)\\ &=\frac{\pi^2}{24} \end{align*}

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On the centralizer

March 24, 2020 / Leave a comment

Suppose that A \in \mathcal{M}_n(\mathbb{C}) has this property that if \lambda is an eigenvalue of A then -\lambda is not an eigenvalue of A. Show that AX=XA if and only if A^2X=XA^2 for any X \in \mathcal{M}_n(\mathbb{C}). In other words the centralizer of A equals the centralizer of A^2.

Solution

It is clear that AX=XA implies A^2X=XA^2 for any X \in \mathcal{M}_n(\mathbb{C}). Now suppose that A^2X=XA^2 for some X \in \mathcal{M}_n(\mathbb{C}) and set Y=AX-XA. We want to prove that Y=0. We have

    \begin{align*}AY+YA &=A(AX-XA)+(AX-XA)A\\ &=A^2X-XA^2\\ &=0 \end{align*}

and so AY=-YA. It now follows that A^kY=(-1)^kYA^k for any integer k \geq 0 and thus for any \lambda \in \mathbb{C} and any integer m \geq 0 we have

    \begin{align*}(A+\lambda \mathbb{I})^mY &=\sum_{k=0}^m\binom{m}{k}\lambda^{m-k}A^kY\\ &=\sum_{k=0}^m\binom{m}{k}(-1)^k\lambda^{m-k}YA^k \\ &=(-1)^mY\sum_{k=0}^m\binom{m}{k}(-\lambda)^{m-k}A^k \\ &=(-1)^mY(A-\lambda \mathbb{I}_{n})^m \quad (*) \end{align*}

where \mathbb{I} is the identity matrix. Now let v be a generalized eigenvector corresponding to an eigenvalue \lambda of A. Then (A-\lambda \mathbb{I}_{n})^mv=0 for some integer m and thus, by (*) we have (A+\lambda \mathbb{I}_n)^mYv=0. Therefore, since we are assuming that -\lambda is not an eigenvalue of A, we must have Yv=0. So, since every element of \mathbb{C}^n is a linear combination of some generalized eigenvectors of A, we get Yu=0 for all u \in \mathbb{C}^n, i.e. Y=0 and hence AX=XA.

The exercise can also be found here.

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Square of a number

March 23, 2020 / Leave a comment

Let a, b, c \in \mathbb{Q} such that \alpha \neq \beta \neq c \neq a. Prove that

    \[\mathcal{A}=\sqrt{\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}}\]

is rational.

Solution

Setting a-b=x ,b-c=y and c-a=z we note that x+y+z=0. Hence,

    \begin{align*} \sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}} &=\sqrt{\frac{x^2y^2+y^2z^2+z^2x^2}{x^2y^2z^2}} \\ &=\sqrt{\frac{(xy+yz+zx)^2-\cancelto{0}{2xyz(x+y+z)}}{x^2y^2z^2}} \\ &=\sqrt{\left(\frac{xy+yz+zx}{xyz}\right)^2} \end{align*}

The result follows.

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A logarithmic integral

March 21, 2020 / 1 Comment on A logarithmic integral

Evaluate the integral

    \[\mathcal{J} = \int_0^1 \left(x \ln x \right)^{2020} \, \mathrm{d}x\]

Solution

Recall the idenity

    \[\mathbf{\ln x = \lim_{n \rightarrow +\infty} n \left ( x^{1/n} - 1 \right )}\]

thus,

    \begin{align*} \int_{0}^{1} \left ( x \ln x \right )^{2020}\, \mathrm{d}x &= \lim_{n \rightarrow +\infty} n^{2020} \int_{0}^{1} \left ( x^{2020} \left ( x^{1/n} -1 \right )^{2020} \right )\, \mathrm{d}x \\ &\!\!\!\!\!\!\overset{u=x^{1/n}}{=\! =\! =\! =\!} \lim_{n \rightarrow +\infty} n^{2021} \int_{0}^{1} u^{2021n-1} \left ( 1-u \right )^{2020} \, \mathrm{d}u \\ &=\lim_{n \rightarrow +\infty} n^{2021} \int_{0}^{1} u^{2021n-1} \left ( 1-u \right )^{2021-1} \, \mathrm{d}u \\ &=\lim_{n \rightarrow +\infty} n^{2021} \mathrm{B} \left ( 2021n, 2021 \right ) \\ &=\lim_{n \rightarrow +\infty} n^{2021} \; \frac{\Gamma \left ( 2021 n \right ) \Gamma \left ( 2021 \right )}{\Gamma \left ( 2021 n + 2021 \right )} \\ &= \Gamma \left ( 2021 \right ) \lim_{n \rightarrow +\infty} n^{2021} \frac{\Gamma \left ( 2021 n \right )}{\Gamma \left ( 2021 n + 2021 \right )} \end{align*}

Using Gautschi’s Inequality it follows that

\displaystyle n^{2021}\left ( 2021n -1 \right )^{1-2022}<\frac{ n^{2021} \Gamma\left ( 2021n -1 + 1 \right )}{\Gamma\left ( 2021n-1 + 2022 \right )}< n^{2021}\left ( 2021n\right )^{1-2022}

and hence the integral equals

    \[\mathcal{J} = \frac{\Gamma(2021)}{2021^{2021}}\]

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