Solution
Solve for \(y\left ( x,t\right ) \) in\begin{equation} y_{tt}=y_{xx}-2\beta y_{t}\qquad \left ( t>0,0<x<\pi \right ) \tag{1} \end{equation} Boundary conditions\begin{align*} y\left ( 0,t\right ) & =0\\ y\left ( \pi ,t\right ) & =0 \end{align*}
Initial conditions\begin{align*} y\left ( x,0\right ) & =f\left ( x\right ) \\ y_{t}\left ( x,0\right ) & =0 \end{align*}
Let \(y=XT\). Substituting in (1) gives\[ T^{\prime \prime }X=X^{\prime \prime }T-2\beta T^{\prime }X \] Dividing by \(XT\neq 0\)\begin{align*} \frac{T^{\prime \prime }}{T} & =\frac{X^{\prime \prime }}{X}-2\beta \frac{T^{\prime }}{T}\\ \frac{T^{\prime \prime }}{T}+2\beta \frac{T^{\prime }}{T} & =\frac{X^{\prime \prime }}{X}=-\lambda \end{align*}
Where \(\lambda \) is separation constant. Due to nature of boundary conditions being both homogeneous, then we know \(\lambda >0\) is only possible case from earlier HW’s. The eigenvalue problem is \[ X^{\prime \prime }+\lambda X=0 \] Which we know has eigenvalues \(\lambda =n^{2}\) for \(n=1,2,\cdots \) with corresponding eigenfunctions \begin{equation} X_{n}=\sin \left ( nx\right ) \tag{1} \end{equation} Now we solve the time ODE using these eigenvalues. \begin{align*} \frac{T^{\prime \prime }}{T}+2\beta \frac{T^{\prime }}{T} & =-n^{2}\\ T^{\prime \prime }+2\beta T^{\prime }+n^{2}T & =0 \end{align*}
This is standard second order ODE with positive damping \(\beta \) and since \(n^{2}\) is positive. The characteristic equation is \[ r^{2}+2\beta r+n^{2}=0 \] The roots are \begin{align*} r & =-\frac{b}{2a}\pm \frac{1}{2a}\sqrt{b^{2}-4ac}\\ & =-\frac{2\beta }{2}\pm \frac{1}{2}\sqrt{4\beta ^{2}-4n^{2}}\\ & =-\beta \pm \sqrt{\beta ^{2}-n^{2}}\\ & =-\beta \pm i\sqrt{n^{2}-\beta ^{2}} \end{align*}
Hence the solution is \begin{align*} T_{n}\left ( t\right ) & =A_{n}e^{r_{1}t}+B_{n}e^{r_{2}t}\\ & =A_{n}e^{\left ( -\beta +i\sqrt{n^{2}-\beta ^{2}}\right ) t}+B_{n}e^{\left ( -\beta -i\sqrt{n^{2}-\beta ^{2}}\right ) t}\\ & =e^{-\beta t}\left ( A_{n}e^{i\sqrt{n^{2}-\beta ^{2}}t}+B_{n}e^{-i\sqrt{n^{2}-\beta ^{2}}t}\right ) \end{align*}
But the above can be rewritten using Euler relation as (the constants \(A_{n},B_{n}\) will be different, but kept them the same names for simplicity)\[ T_{n}\left ( t\right ) =e^{-\beta t}\left ( A_{n}\cos \left ( \sqrt{n^{2}-\beta ^{2}}t\right ) +B_{n}\sin \left ( \sqrt{n^{2}-\beta ^{2}}t\right ) \right ) \] Let \(\alpha _{n}=\sqrt{n^{2}-\beta ^{2}}\), then the above becomes\begin{equation} T_{n}\left ( t\right ) =e^{-\beta t}\left ( A_{n}\cos \left ( \alpha _{n}t\right ) +B_{n}\sin \left ( \alpha _{n}t\right ) \right ) \tag{2} \end{equation} Since the PDE is linear and homogenous, then by superposition we obtain the final solution as\begin{align} y\left ( x,t\right ) & =\sum _{n=1}^{\infty }X_{n}T_{n}\nonumber \\ & =\sum _{n=1}^{\infty }e^{-\beta t}\left ( A_{n}\cos \left ( \alpha _{n}t\right ) +B_{n}\sin \left ( \alpha _{n}t\right ) \right ) \sin \left ( nx\right ) \tag{3} \end{align}
Now initial conditions are applied to determine \(A_{n},B_{n}\). At \(t=0\)\[ f\left ( x\right ) =\sum _{n=1}^{\infty }A_{n}\sin \left ( nx\right ) \] Hence \(A_{n}\) are the Fourier sine coefficient of the representation of \(f\left ( x\right ) \) which implies\begin{equation} A_{n}=\frac{2}{\pi }\int _{0}^{\pi }f\left ( x\right ) \sin \left ( nx\right ) dx\tag{4} \end{equation} Taking time derivative of (3) gives\[ y_{t}\left ( x,t\right ) =\sum _{n=1}^{\infty }\left [ -\beta e^{-\beta t}\left ( A_{n}\cos \left ( \alpha _{n}t\right ) +B_{n}\sin \left ( \alpha _{n}t\right ) \right ) +e^{-\beta t}\left ( -\alpha _{n}A_{n}\sin \left ( \alpha _{n}t\right ) +\alpha _{n}B_{n}\cos \left ( \alpha _{n}t\right ) \right ) \right ] \sin \left ( nx\right ) \] At \(t=0\) the above becomes (since released from rest)\[ 0=\sum _{n=1}^{\infty }\left ( -\beta A_{n}+\alpha _{n}B_{n}\right ) \sin \left ( nx\right ) \] Therefore \[ -\beta A_{n}+\alpha _{n}B_{n}=0 \] Hence \(B_{n}=\frac{\beta A_{n}}{\alpha _{n}}\). Therefore (3) becomes\begin{align*} y\left ( x,t\right ) & =\sum _{n=1}^{\infty }e^{-\beta t}\left ( A_{n}\cos \left ( \alpha _{n}t\right ) +\frac{\beta A_{n}}{\alpha _{n}}\sin \left ( \alpha _{n}t\right ) \right ) \sin \left ( nx\right ) \\ & =e^{-\beta t}\sum _{n=1}^{\infty }A_{n}\left ( \cos \left ( \alpha _{n}t\right ) +\frac{\beta }{\alpha _{n}}\sin \left ( \alpha _{n}t\right ) \right ) \sin \left ( nx\right ) \end{align*}
Where \(A_{n}=\frac{2}{\pi }\int _{0}^{\pi }f\left ( x\right ) \sin \left ( nx\right ) dx\) and \(\alpha _{n}=\sqrt{n^{2}-\beta ^{2}}\). Which is the result required to show (Book used \(B\) in place \(A\), but it is the same thing, just different name for a constant).
Solution
suppose \(\omega \neq a\). Let \begin{equation} y_{p}=A\cos \omega t+B\sin \omega t \tag{1} \end{equation} Then\begin{align*} y_{p}^{\prime } & =-A\omega \sin \omega t+B\omega \cos \omega t\\ y_{p}^{\prime \prime } & =-A\omega ^{2}\cos \omega t-B\omega ^{2}\sin \omega t \end{align*}
Substituting the above back into the given ODE gives\begin{align} y_{p}^{\prime \prime }\left ( t\right ) +a^{2}y_{p}\left ( t\right ) & =b\sin \omega t\nonumber \\ \left ( -A\omega ^{2}\cos \omega t-B\omega ^{2}\sin \omega t\right ) +a^{2}\left ( A\cos \omega t+B\sin \omega t\right ) & =b\sin \omega t\nonumber \\ \cos \omega t\left ( -A\omega ^{2}+a^{2}A\right ) +\sin \omega t\left ( -B\omega ^{2}+a^{2}B\right ) & =b\sin \omega t \tag{2} \end{align}
By comparing coefficients, we see that\begin{align*} -A\omega ^{2}+a^{2}A & =0\\ A\left ( a^{2}-\omega ^{2}\right ) & =0 \end{align*}
Since \(\omega \neq a\) then this implies that \(A=0\). And from (2), we see that\begin{align*} -B\omega ^{2}+a^{2}B & =b\\ B & =\frac{b}{a^{2}-\omega ^{2}} \end{align*}
Therefore (1) becomes\begin{equation} y_{p}=\frac{b}{a^{2}-\omega ^{2}}\sin \omega t \tag{3} \end{equation} Now we need to find the complementary solution to \[ y_{c}^{\prime \prime }+a^{2}y=0 \] Since \(a^{2}>0\), then the solution is the standard one given by\begin{equation} y_{c}\left ( t\right ) =C_{1}\cos at+C_{2}\sin at \tag{4} \end{equation} Adding (3,4) gives the general solution\[ y\left ( t\right ) =C_{1}\cos at+C_{2}\sin at+\frac{b}{a^{2}-\omega ^{2}}\sin \omega t \]
Let \begin{equation} y_{p}=At\cos \omega t+Bt\sin \omega t \tag{1} \end{equation} Then\begin{align*} y_{p}^{\prime } & =A\cos \omega t-At\omega \sin \omega t+B\sin \omega t+Bt\omega \cos \omega t\\ y_{p}^{\prime \prime } & =-A\omega \sin \omega t-\left ( A\omega \sin \omega t+At\omega ^{2}\cos \omega t\right ) +B\omega \cos \omega t+\left ( B\omega \cos \omega t-Bt\omega ^{2}\sin \omega t\right ) \\ & =\left ( -At\omega ^{2}+2B\omega \right ) \cos \omega t+\left ( -2A\omega -Bt\omega ^{2}\right ) \sin \omega t \end{align*}
Substituting the above back into the given ODE gives\begin{align} y_{p}^{\prime \prime }\left ( t\right ) +a^{2}y_{p}\left ( t\right ) & =b\sin \omega t\nonumber \\ \left ( \left ( -At\omega ^{2}+2B\omega \right ) \cos \omega t+\left ( -2A\omega -Bt\omega ^{2}\right ) \sin \omega t\right ) +a^{2}\left ( At\cos \omega t+Bt\sin \omega t\right ) & =b\sin \omega t\nonumber \\ \cos \omega t\left ( -At\omega ^{2}+2B\omega +a^{2}At\right ) +\sin \omega t\left ( -2A\omega -Bt\omega ^{2}+a^{2}Bt\right ) & =b\sin \omega t \tag{2} \end{align}
By comparing coefficients, we see that\begin{align} -At\omega ^{2}+2B\omega +a^{2}At & =0\nonumber \\ At\left ( -\omega ^{2}+a^{2}\right ) +B\left ( 2\omega \right ) & =0 \tag{3} \end{align}
And from (2), we see also that\begin{align} -2A\omega -Bt\omega ^{2}+a^{2}Bt & =b\nonumber \\ A\left ( -2\omega \right ) +Bt\left ( -\omega ^{2}+a^{2}\right ) & =b \tag{4} \end{align}
But since \(\omega =a\), then (3) becomes\begin{align*} B\left ( 2\omega \right ) & =0\\ B & =0 \end{align*}
And (4) becomes\begin{align*} A\left ( -2\omega \right ) & =b\\ A & =\frac{-b}{2a} \end{align*}
Substituting these values we found for \(A,B\), in (1) gives\[ y_{p}=\frac{-b}{2a}t\cos \omega t \] But \(\omega =a\), therefore\begin{equation} y_{p}=\frac{-b}{2a}t\cos at \tag{5} \end{equation} The complementary solution do not change from part (a). Hence the general solution is\[ y\left ( t\right ) =C_{1}\cos at+C_{2}\sin at-\frac{b}{2a}t\cos at \] Which is the result required to show.
Solution
The general solution from problem 2 is\[ y\left ( t\right ) =\left \{ \begin{array} [c]{cc}C_{1}\cos at+C_{2}\sin at+\frac{b}{a^{2}-\omega ^{2}}\sin \omega t & \omega \neq a\\ C_{1}\cos at+C_{2}\sin at-\frac{b}{2a}t\cos at & \omega =a \end{array} \right . \] We need to find \(C_{1},C_{2}\) when initial conditions are \(y\left ( 0\right ) =0,y^{\prime }\left ( 0\right ) =0\) for each of the above cases.
case \(\omega \neq a\)
\(y\left ( 0\right ) =0\) gives\[ 0=C_{1}\] Hence solution now becomes\[ y\left ( t\right ) =C_{2}\sin at+\frac{b}{a^{2}-\omega ^{2}}\sin \omega t \] Taking time derivative gives\[ y^{\prime }\left ( t\right ) =aC_{2}\cos at+\frac{\omega b}{a^{2}-\omega ^{2}}\cos \omega t \] At \(t=0\) the above gives\begin{align*} 0 & =aC_{2}+\frac{\omega b}{a^{2}-\omega ^{2}}\\ C_{2} & =\frac{1}{a}\frac{\omega b}{\omega ^{2}-a^{2}} \end{align*}
Using \(C_{1},C_{2}\) found above, the solution becomes\begin{align} y\left ( t\right ) & =\frac{1}{a}\frac{\omega b}{\omega ^{2}-a^{2}}\sin at+\frac{b}{a^{2}-\omega ^{2}}\sin \omega t\nonumber \\ & =\frac{b}{a^{2}-\omega ^{2}}\left ( \frac{\omega }{a}\sin at-\sin \omega t\right ) \tag{1} \end{align}
case \(\omega =a\)
\(y\left ( 0\right ) =0\) gives\[ 0=C_{1}\] Hence solution now becomes\[ y\left ( t\right ) =C_{2}\sin at-\frac{b}{2a}t\cos at \] Taking time derivative gives\[ y^{\prime }\left ( t\right ) =aC_{2}\cos at-\left ( \frac{b}{2a}\cos at-\frac{b}{2a}t^{2}\sin at\right ) \] At \(t=0\) the above gives\begin{align*} 0 & =aC_{2}-\frac{b}{2a}\\ C_{2} & =\frac{1}{a}\frac{b}{2a} \end{align*}
Using \(C_{1},C_{2}\) found above, the solution becomes\begin{align} y\left ( t\right ) & =\frac{1}{a}\frac{b}{2a}\sin at-\frac{b}{2a}t\cos at\nonumber \\ & =\frac{b}{2a}\left ( \frac{1}{a}\sin at-t\cos at\right ) \tag{2} \end{align}
From (1,2) we see that\[ y\left ( t\right ) =\left \{ \begin{array} [c]{cc}\frac{b}{a^{2}-\omega ^{2}}\left ( \frac{\omega }{a}\sin at-\sin \omega t\right ) & \omega \neq a\\ \frac{b}{2a}\left ( \frac{1}{a}\sin at-t\cos at\right ) & \omega =a \end{array} \right . \] Which is the result required to show.
Solution
\begin{align*} f\left ( x\right ) & =\int _{0}^{\infty }\left ( A\left ( \alpha \right ) \cos \left ( \alpha x\right ) +B\left ( \alpha \right ) \sin \left ( \alpha x\right ) \right ) d\alpha \\ & =\int _{0}^{\infty }\left ( A\left ( \alpha \right ) \left ( \frac{e^{i\alpha x}+e^{-i\alpha x}}{2}\right ) -iB\left ( \alpha \right ) \left ( \frac{e^{i\alpha x}-e^{-i\alpha x}}{2}\right ) \right ) d\alpha \\ & =\int _{0}^{\infty }\left ( e^{i\alpha x}\left ( \frac{A\left ( \alpha \right ) -iB\left ( \alpha \right ) }{2}\right ) +e^{-i\alpha x}\left ( \frac{A\left ( \alpha \right ) +iB\left ( \alpha \right ) }{2}\right ) \right ) d\alpha \\ & =\int _{0}^{\infty }e^{i\alpha x}\frac{A\left ( \alpha \right ) -iB\left ( \alpha \right ) }{2}d\alpha +\int _{0}^{\infty }e^{-i\alpha x}\frac{A\left ( \alpha \right ) +iB\left ( \alpha \right ) }{2}d\alpha \\ & =\int _{0}^{\infty }e^{i\alpha x}\frac{A\left ( \alpha \right ) -iB\left ( \alpha \right ) }{2}d\alpha +\int _{0}^{\infty }e^{-i\alpha x}\frac{A\left ( \alpha \right ) +iB\left ( \alpha \right ) }{2}d\alpha \\ & =\int _{0}^{\infty }e^{i\alpha x}\frac{A\left ( \alpha \right ) -iB\left ( \alpha \right ) }{2}d\alpha +\int _{-\infty }^{0}e^{i\alpha x}\frac{A\left ( \alpha \right ) +iB\left ( \alpha \right ) }{2}d\alpha \\ & =\int _{-\infty }^{\infty }C\left ( \alpha \right ) e^{i\alpha x}d\alpha \end{align*}
Where \[\begin{array} [c]{ccc}C\left ( \alpha \right ) =\frac{A\left ( \alpha \right ) -iB\left ( \alpha \right ) }{2} & , & C\left ( -\alpha \right ) =\frac{A\left ( \alpha \right ) +iB\left ( \alpha \right ) }{2}\end{array} \qquad \alpha >0 \] Expression (9) section (5) is\begin{align*} A\left ( \alpha \right ) & =\frac{1}{\pi }\int _{-\infty }^{\infty }f\left ( x\right ) \cos \left ( \alpha x\right ) dx\\ B\left ( \alpha \right ) & =\frac{1}{\pi }\int _{-\infty }^{\infty }f\left ( x\right ) \sin \left ( \alpha x\right ) dx \end{align*}
Substituting the above in \(C\left ( \alpha \right ) =\frac{A\left ( \alpha \right ) -iB\left ( \alpha \right ) }{2}\) gives\begin{align*} C\left ( \alpha \right ) & =\frac{1}{2}\left ( \frac{1}{\pi }\int _{-\infty }^{\infty }f\left ( x\right ) \cos \left ( \alpha x\right ) dx-i\frac{1}{\pi }\int _{-\infty }^{\infty }f\left ( x\right ) \sin \left ( \alpha x\right ) dx\right ) \\ & =\frac{1}{2\pi }\left ( \int _{-\infty }^{\infty }f\left ( x\right ) \cos \left ( \alpha x\right ) dx-\int _{-\infty }^{\infty }f\left ( x\right ) i\sin \left ( \alpha x\right ) dx\right ) \\ & =\frac{1}{2\pi }\int _{-\infty }^{\infty }f\left ( x\right ) \left ( \cos \left ( \alpha x\right ) -i\sin \left ( \alpha x\right ) \right ) dx \end{align*}
But using Euler relation \(\cos \left ( \alpha x\right ) -i\sin \left ( \alpha x\right ) =e^{i\alpha x}\) then the above reduces to\[ C\left ( \alpha \right ) =\frac{1}{2\pi }\int _{-\infty }^{\infty }f\left ( x\right ) e^{i\alpha x}dx\qquad -\infty <\alpha <\infty \] Which is what required to show.
Solution
Since \(f\left ( x\right ) \) is piecewise continuous and absolutely integrable (sine function), then\[ \frac{f\left ( x^{+}\right ) +f\left ( x^{-}\right ) }{2}=\frac{1}{\pi }\int _{0}^{\infty }\left ( \int _{-\infty }^{\infty }f\left ( s\right ) \cos \left ( \alpha \left ( s-x\right ) \right ) ds\right ) d\alpha \] Substituting for \(f\left ( s\right ) \) inside the integral for the function given gives\[ \frac{f\left ( x^{+}\right ) +f\left ( x^{-}\right ) }{2}=\frac{1}{\pi }\int _{0}^{\infty }\left ( \int _{0}^{\pi }\sin \left ( s\right ) \cos \left ( \alpha s-\alpha x\right ) ds\right ) d\alpha \] Where we used \(\int _{0}^{\pi }\) only, since the function is zero everywhere else. Using \(2\sin A\cos B=\sin \left ( A+B\right ) +\sin \left ( A-B\right ) \) then the above can be written as\begin{align} \frac{f\left ( x^{+}\right ) +f\left ( x^{-}\right ) }{2} & =\frac{1}{\pi }\int _{0}^{\infty }\left ( \frac{1}{2}\int _{0}^{\pi }\sin \left ( s+\alpha s-\alpha x\right ) +\sin \left ( s-\left ( \alpha s-\alpha x\right ) \right ) ds\right ) d\alpha \nonumber \\ & =\frac{1}{2\pi }\int _{0}^{\infty }\left ( \int _{0}^{\pi }\sin \left ( s+\alpha s-\alpha x\right ) +\sin \left ( s-\alpha s+\alpha x\right ) ds\right ) d\alpha \tag{1} \end{align}
But \begin{align*} \int _{0}^{\pi }\sin \left ( s+\alpha s-\alpha x\right ) ds & =\left [ \frac{-\cos \left ( s+\alpha s-\alpha x\right ) }{1+\alpha }\right ] _{0}^{\pi }\\ & =\frac{-1}{1+\alpha }\left ( \cos \left ( \pi +\alpha \pi -\alpha x\right ) -\cos \left ( -\alpha x\right ) \right ) \\ & =\frac{-1}{1+\alpha }\left ( \cos \left ( \pi +\alpha \left ( \pi -x\right ) \right ) -\cos \left ( \alpha x\right ) \right ) \end{align*}
But \(\cos \left ( \pi +\alpha \left ( \pi -x\right ) \right ) =-\cos \left ( \alpha \left ( \pi -x\right ) \right ) \), and the above becomes\begin{equation} \int _{0}^{\pi }\sin \left ( s+\alpha s-\alpha x\right ) ds=\frac{1}{1+\alpha }\left ( \cos \left ( \alpha \left ( \pi -x\right ) \right ) +\cos \left ( \alpha x\right ) \right ) \tag{2} \end{equation} Similarly \begin{align} \int _{0}^{\pi }\sin \left ( s-\alpha s+\alpha x\right ) ds & =\left [ \frac{-\cos \left ( s-\alpha s+\alpha x\right ) }{1-\alpha }\right ] _{0}^{\pi }\nonumber \\ & =\frac{-1}{1-\alpha }\left ( \cos \left ( \pi -\alpha \pi +\alpha x\right ) -\cos \left ( \alpha x\right ) \right ) \nonumber \\ & =\frac{-1}{1-\alpha }\left ( \cos \left ( \pi -\alpha \left ( \pi +x\right ) \right ) -\cos \left ( \alpha x\right ) \right ) \nonumber \\ & =\frac{-1}{1-\alpha }\left ( -\cos \left ( -\alpha \left ( \pi +x\right ) \right ) -\cos \left ( \alpha x\right ) \right ) \nonumber \\ & =\frac{1}{1-\alpha }\left ( \cos \left ( \alpha \left ( \pi +x\right ) \right ) +\cos \left ( \alpha x\right ) \right ) \tag{3} \end{align}
Substituting (2,3) back in (1) gives\begin{align*} \frac{f\left ( x^{+}\right ) +f\left ( x^{-}\right ) }{2} & =\frac{1}{2\pi }\int _{0}^{\infty }\left ( \frac{1}{1+\alpha }\left ( \cos \left ( \alpha \left ( \pi -x\right ) \right ) +\cos \left ( \alpha x\right ) \right ) +\frac{1}{1-\alpha }\left ( \cos \left ( \alpha \left ( \pi +x\right ) \right ) +\cos \left ( \alpha x\right ) \right ) \right ) d\alpha \\ & =\frac{1}{2\pi }\int _{0}^{\infty }\left ( \cos \left ( \alpha \left ( \pi -x\right ) \right ) \left ( \frac{1}{1+\alpha }+\frac{1}{1-\alpha }\right ) +\cos \left ( \alpha x\right ) \left ( \frac{1}{1+\alpha }+\frac{1}{1-\alpha }\right ) \right ) d\alpha \\ & =\frac{1}{2\pi }\int _{0}^{\infty }\left ( \cos \left ( \alpha \left ( \pi -x\right ) \right ) \left ( \frac{2}{1-\alpha ^{2}}\right ) +\cos \left ( \alpha x\right ) \left ( \frac{2}{1-\alpha ^{2}}\right ) \right ) d\alpha \\ & =\frac{1}{\pi }\int _{0}^{\infty }\frac{\cos \left ( \alpha \left ( \pi -x\right ) \right ) +\cos \left ( \alpha x\right ) }{1-\alpha ^{2}}d\alpha \end{align*}
But \(f\left ( x\right ) \) is continuous then \(\frac{f\left ( x^{+}\right ) +f\left ( x^{-}\right ) }{2}=f\left ( x\right ) \) and the above becomes\[ f\left ( x\right ) =\frac{1}{\pi }\int _{0}^{\infty }\frac{\cos \left ( \alpha \left ( \pi -x\right ) \right ) +\cos \left ( \alpha x\right ) }{1-\alpha ^{2}}d\alpha \] When \(x=\frac{\pi }{2}\) the above gives \[ f\left ( \frac{\pi }{2}\right ) =\frac{1}{\pi }\int _{0}^{\infty }\frac{\cos \left ( \alpha \left ( \pi -\frac{\pi }{2}\right ) \right ) +\cos \left ( \alpha \frac{\pi }{2}\right ) }{1-\alpha ^{2}}d\alpha \] But \(f\left ( \frac{\pi }{2}\right ) =\sin \left ( \frac{\pi }{2}\right ) =1\), hence\begin{align*} 1 & =\frac{1}{\pi }\int _{0}^{\infty }\frac{\cos \left ( \alpha \frac{\pi }{2}\right ) +\cos \left ( \alpha \frac{\pi }{2}\right ) }{1-\alpha ^{2}}d\alpha \\ & =\frac{1}{\pi }\int _{0}^{\infty }\frac{2\cos \left ( \alpha \frac{\pi }{2}\right ) }{1-\alpha ^{2}}d\alpha \end{align*}
Therefore\[ \frac{\pi }{2}=\int _{0}^{\infty }\frac{\cos \left ( \alpha \frac{\pi }{2}\right ) }{1-\alpha ^{2}}d\alpha \]