Find $f(x)$ from $\int_0^\infty f(x)\cos(\alpha x)\,dx=\frac{e^{-\alpha}\sin(\alpha)}{\alpha}$

Question

Consider the following integration equation: $\int_{0}^{\infty}f(x) \cos\left(\alpha x\right)dx=\frac{e^{-\alpha}\sin\left(\alpha\right)}{\alpha}$

I want to compute $f(x)$. For this, I use Fourier integration. But the results missed. Please help me.

Answer 1

Since

$$\int_0^{\infty} e^{-x}\cos(\alpha x)\ dx = \frac{1}{\alpha^2+1}$$

we can take

$$f(x) = e^{-\alpha-x}\left(\alpha+\frac{1}{\alpha}\right)\sin(\alpha)$$

which works.

Answer 2

We will restrict ourselves to $\xi>0$. Let us extend $f$ even across the $y$-axis to a symmetric function $f(x) = f(-x)$. Then we see \begin{align} \sqrt{2\pi}\hat f(\xi)=&\ \int^\infty_{-\infty} e^{- i x\xi} f(x)\ dx = \int^\infty_0 e^{-i x\xi}f(x)\ dx + \int^0_{-\infty} e^{- i x\xi}f(x)\ dx\\ =&\ 2\int^\infty_0 \cos(x\xi) f(x)\ dx \end{align} which means we are trying to find an even function $f$ such that \begin{align} \int^\infty_0 \cos(x\xi) f(x)\ dx = \sqrt{\frac{\pi}{2}} \hat f(\xi) = \frac{e^{-\xi}\sin \xi}{\xi} \end{align} for all $\xi>0$ or equivalently \begin{align} \hat f(\xi) = \sqrt{\frac{2}{\pi}}\frac{e^{-\xi}\sin \xi}{\xi} \end{align} for $\xi>0$. If we extend symmetric $\hat f(\xi)$ about $\xi=0$, then we are interested in finding $f$ such that \begin{align} \hat f(\xi) = \sqrt{\frac{2}{\pi}}\frac{e^{-|\xi|}\sin \xi}{\xi}= \sqrt{\frac{2}{\pi}} e^{-|\xi|}\cdot \frac{\sin \xi}{\xi} \end{align} which means \begin{align} f(x) = \sqrt{\frac{2}{\pi}}(e^{-|\xi|})^\vee \ast (\operatorname{sinc}(\xi))^\vee(x). \end{align} Since we have following inversion \begin{align} (\operatorname{sinc}(\xi))^\vee(x) = \chi_{[-1, 1]}(x) \end{align} and \begin{align} (e^{-|\xi|})^\vee(x) = \sqrt{\frac{2}{\pi}} \frac{1}{1+x^2} \end{align} which means \begin{align} f(x) =&\ \frac{2}{\pi}\int^\infty_{-\infty} \frac{1}{1+(x-y)^2} \chi_{[-1, 1]}(y)\ dy =\frac{2}{\pi}\int^1_{-1} \frac{1}{1+(x-y)^2}\ dy \\ =&\ \frac{2}{\pi} [\arctan(x+1)-\arctan(x-1)]. \end{align}

Remark: The constant $2/\pi$ might be off.

Answer 3

The answer is the inverse Fourier cosine transform: \begin{align} f(x) & = \frac{2}{\pi}\int_{0}^{\infty}e^{-\alpha}\frac{\sin\alpha}{\alpha}\cos(\alpha x)d\alpha, \\ f'(x)& =-\frac{2}{\pi}\int_{0}^{\infty}e^{-\alpha}\sin(\alpha)\sin(\alpha x)d\alpha \\ &=\frac{1}{\pi}\int_{0}^{\infty}e^{-\alpha}\{\cos(\alpha(x+1))-\cos(\alpha(x-1))\}d\alpha \\ &=\Re\frac{1}{\pi}\int_{0}^{\infty}e^{-\alpha(1+i(x+1))}-e^{-\alpha(1+i(x-1))}d\alpha \\ &=\Re\frac{1}{\pi}\left[\frac{1}{1+i(x+1)}-\frac{1}{1+i(x-1)}\right] \\ &=\frac{1}{\pi}\left[\frac{1}{1+(x+1)^2}-\frac{1}{1+(x-1)^2}\right] \\ f(x) &= \frac{1}{\pi}\{\tan^{-1}(1+x)-\tan^{-1}(x-1)\}+C. \end{align} Evaluating the original expression at $x=0$ gives $C$: $$ \frac{2}{\pi}\int_{0}^{\infty}e^{-\alpha}\frac{\sin\alpha}{\alpha}d\alpha = \frac{2}{\pi}2\tan^{-1}(1)+C = 1+C. $$