Find the Fourier sine integal for $f(x)= e^{-bx}$ and prove that
$$\frac{\pi}{2}f(x)=\int_0^{\infty}\frac{t\sin(tx)}{b^2+t^2}dt.$$
I don't want the solution to the problem, I was able to solve it. Instead I want to know why can we use integral Fourier sine integral for this $f(x)$ as it can be used for $f(x)$ being odd only. But here $f(x)$ is neither even nor odd. The sine functions may not form the complete basis. Please illuminate on this.
Consider the function
$$f_b(x)=\text{sgn}(x)\ e^{-b\,|x|}\,,\quad \Re(b)>0\tag{1}$$
which is an odd function of $x$ as illustrated in Figure (1) below.
Figure (1): Illustration of $f_1(x)$
Since $f_b(x)$ is an odd function of $x$, the Fourier transform of $f_b(x)$ is simply $i$ times the Fourier sine transform of $f_b(x)$ as illustrated in formula (2) below
$$\mathcal{F}_x[f_b(x)](t)=i\ \text{FourierSinTransform}[f_b(x),x,t]=\frac{i \sqrt{\frac{2}{\pi }}\ t}{b^2+t^2}\,,\quad \Re(b)>0\tag{2}$$
and
$$\text{FourierSinTransform}[f_b(x),x,t]=\text{FourierSinTransform}\left[e^{-b\,x},x,t\right]=\frac{\sqrt{\frac{2}{\pi}}\ t}{b^2+t^2}\,,\quad \Re(b)>0\tag{3}$$
Now consider the function
$$g_b(x)=e^{-b\,|x|}\,,\quad \Re(b)>0\tag{4}$$
which is an even function of $x$ as illustrated in Figure (2) below.
Figure (2): Illustration of $g_1(x)$
Since $g_b(x)$ is an even function of $x$, the Fourier transform of $g_b(x)$ is simply the Fourier cosine transform of $g_b(x)$ as illustrated in formula (4) below
$$\mathcal{F}_x[g(x,b)](t)=\text{FourierCosTransform}[g(x,b),x,t]=\frac{\sqrt{\frac{2}{\pi}}\ b}{b^2+t^2}\,,\quad \Re(b)>0\tag{5}$$
and
$$\text{FourierCosTransform}[g(x,b),x,t]=\text{FourierCosTransform}\left[e^{-b\,x},x,t\right]=\frac{\sqrt{\frac{2}{\pi}}\ b}{b^2+t^2}\,,\quad \Re(b)>0\tag{6}$$
Note $f_b(x)=e^{-b\,x}$ is only valid for $x>0$, and $g_b(x)=e^{-b\,x}$ is only valid for $x\ge 0$.
The results above assume the following definitions of the Fourier, Fourier sine, and Fourier cosine transforms.
$$\mathcal{F}_x[f(x)](t)=\frac{1}{\sqrt{2 \pi}}\int\limits_{-\infty}^\infty f(x)\ e^{i\,t\,x}\,dx\tag{7}$$
$$\text{FourierSinTransform}[f(x),x,t]=\sqrt{\frac{2}{\pi}}\int\limits_0^\infty f(x) \sin(t\,x)\,dx\tag{8}$$
$$\text{FourierCosTransform}[f(x),x,t]=\sqrt{\frac{2}{\pi}}\int\limits_0^\infty f(x) \cos(t\,x)\,dx\tag{8}$$