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Trigonometric integral

Prove that

    \[\bigintsss_0^{\infty}\frac{\sin \sqrt{x^2+1} \cos x}{\sqrt{x^2+1}} \, \mathrm{d}x=\frac{\pi}{4}\]

Solution

Let \mathcal{J} be the integral. Note that

    \[2 \sin \sqrt{x^2+1} \cos x =  \sin \left ( \sqrt{1+x^2} - x \right ) + \sin \left ( \sqrt{1+x^2}+x \right ) \]

and hence:

    \[2\mathcal{J} = \bigintsss_{0}^{\infty} \frac{\sin \left ( \sqrt{1+x^2}-x \right )}{\sqrt{1+x^2}} \,\mathrm{d}x + \bigintsss_{0}^{\infty} \frac{\sin \left ( \sqrt{1+x^2} + x \right )}{\sqrt{1+x^2}} \, \mathrm{d}x\]

For the integral \bigintsss_{0}^{\infty} \frac{\sin \left ( \sqrt{1+x^2}-x \right )}{\sqrt{1+x^2}} \,\mathrm{d}x we apply the substitution t \mapsto \sqrt{1+x^2}-x. Then, x = \frac{1-t^2}{2t} and

(1)   \begin{equation*} \bigintsss_{0}^{\infty} \frac{\sin \left ( \sqrt{1+x^2}-x \right )}{\sqrt{1+x^2}} \,\mathrm{d}x = \int_{0}^{1} \frac{\sin t}{t} \, \mathrm{d}t \end{equation*}

and similarly by applying the change of variables t \mapsto \sqrt{1+x^2} + x at the second integral we get that

(2)   \begin{equation*} \int_{0}^{\infty} \frac{\sin \left ( \sqrt{1+x^2} \right )}{\sqrt{1+x^2}} \, \mathrm{d}x = \int_{1}^{\infty} \frac{\sin t}{t} \, \mathrm{d}t \end{equation*}

Adding equations (1) , (2) we get that

    \begin{align*} 2\mathcal{J} &= \int_{0}^{1} \frac{\sin t}{t} \, \mathrm{d}t + \int_{1}^{\infty} \frac{\sin t}{t} \, \mathrm{d}t \\ &= \int_{0}^{\infty} \frac{\sin t}{t} \, \mathrm{d}t \\ &= \frac{\pi}{2} \end{align*}

and the result follows.

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Parametric integral

Let a, b>0. Prove that:

    \[\int_{a}^{b} \frac{\mathrm{d}t}{t\sqrt{\left ( t-a \right )\left (b -t  \right )} } = \frac{\pi}{\sqrt{ab}}\]

Solution

We’re applying the change of variables t \mapsto a \cos^2 \theta + b \sin^2 \theta and thus,

\begin{aligned} \int_{a}^{b} \frac{\mathrm{d}t}{t\sqrt{\left ( t-a \right )\left (b -t \right )}} &=\int_{0}^{\pi/2} \frac{2\left ( b-a \right )\sin \theta \cos \theta}{\left ( a \cos^2 \theta + b \sin^2 \theta \right ) \sqrt{\left ( b-a \right )^2 \sin^2 \theta \cos^2 \theta}} \, \mathrm{d}\theta \\ &=2 \int_{0}^{\pi/2} \frac{\mathrm{d} \theta}{a \cos^2 \theta + b \sin^2 \theta} \\ &=2 \int_{0}^{\pi/2} \frac{\mathrm{d}\theta}{\cos^2 \theta \left ( a + b \tan^2 \theta \right )} \\ &=2 \int_{0}^{\pi/2} \frac{\sec^2 \theta}{a + b \tan^2 \theta } \, \mathrm{d} \theta \\ &\!\!\!\!\!\overset{y = \tan \theta}{=\! =\! =\! =\! =\!} 2 \int_{0}^{\infty} \frac{\mathrm{d}y}{a + b y^2} \\ &= \frac{\pi}{\sqrt{ab}} \end{aligned}

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Abel’s Integral

Prove that:

    \[\int_{0}^{\infty} \frac{t}{\left ( e^{\pi t} -e^{-\pi t} \right )\left ( 1+t^2 \right )}\, \mathrm{d}t = \frac{\ln 2}{2} - \frac{1}{4}\]

Solution

We are applying Abel – Plana. We choose f(t)=\frac{1}{2(1+t)} thus,

\begin{aligned} i \int_{0}^{\infty} \frac{f\left ( it \right )- f\left ( -it \right )}{2 \sinh \pi t}\, \mathrm{d}t + \frac{f(0)}{2} = \sum_{n=0}^{\infty} (-1)^n f(n) &\Leftrightarrow i \int_{0}^{\infty} \frac{\frac{1}{2\left ( 1+it \right )}- \frac{1}{2\left ( 1-it \right )}}{2 \sinh \pi t}\, \mathrm{d}t + \frac{1}{4} = \\ &\quad \quad \quad \quad = \sum_{n=0}^{\infty} \frac{(-1)^n}{2\left ( 1+n \right )}\\ &\Leftrightarrow -i^2 \int_{0}^{\infty} \frac{t}{2\sinh \pi t \left ( t^2+1 \right )} \, \mathrm{d}t + \frac{1}{4} = \\ &\quad \quad \quad \quad = \sum_{n=0}^{\infty} \frac{(-1)^n}{2\left ( 1+n \right )} \\ &\Leftrightarrow \int_{0}^{\infty} \frac{t}{2\sinh \pi t \left ( t^2+1 \right )} \, \mathrm{d}t + \frac{1}{4} = \frac{\ln 2}{2} \\ &\Leftrightarrow \int_{0}^{\infty} \frac{t}{2\sinh \pi t \left ( t^2+1 \right )} \, \mathrm{d}t = \frac{\ln 2}{2} - \frac{1}{4} \\ &\Leftrightarrow \int_{0}^{\infty} \frac{t}{\left ( e^{\pi t} -e^{-\pi t} \right )\left ( 1+t^2 \right )}\, \mathrm{d}t = \frac{\ln 2}{2} - \frac{1}{4} \end{aligned}

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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

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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