A problem about Joint sine and cosine fourier transform

There is a problem on Page 60 of Book of B.M.Budak, A. A. Samarskii and A. N. Tikhonov, whose title is A Collection of Problems in Mathematical Physics (New York, Dover, 1964). The problem (the problem number is 64) is: Applying the Fourier transform with the kernel $$K(x,\lambda)=\sqrt{\frac{2}{\pi}}\frac{\lambda \cos (\lambda x)+h\sin(\lambda x)}{\lambda^2+h^2},$$ solve the boundary-value problem

$u_t=a^2 u_{xx},$ $0< x,t< +\infty$, $u_x(0,t)-hu(0,t)=-h\phi(t), t>0$, $u(x,0)=0,0< x<+\infty$.

Apparently, this transform is just the joint sine and cosine transform. But I have never seen this kind of Fourier transform before, and therefore I do not how to use this kind of integral transform to solve the problem here, although I've tried several days. And I do not even know what is its inverse transform:

If we denote the image of $f(x)$ of this transform by $\hat u(\lambda)=\int_0^{+\infty} K(x,\lambda)u(x)dx,$ is it true that $u(x)=\int_0^{+\infty} K(x,\lambda)\hat{u}(\lambda)d\lambda$ ?

I could not find any other books about this kind of Fourier transform. Can anyone help me solve this problem? or just give me some hints on the properties of the Fourier transform with the above kernel $K(x,\lambda)$?

• hmm, there is no square root in the denominator? if there was a square root, then we could define $\alpha=\arccos(\lambda/\sqrt{\lambda^2+h^2})$ and your kernel would simply be $K(x,\lambda)=\sqrt{2/\pi}\cos(\lambda x-\alpha)$ – Carlo Beenakker May 11 '13 at 19:36
• There is indeed no square root in the denominator of the kernel $K(x,\lambda)$. I have tried the method just as you thought, but the result is no use. – azhi May 12 '13 at 0:51
• The thing is this is just a solution to Sturm-Liouville problem on an infinite interval, the eigenvalues will form continuous spectrum, so the Fourier series will be replaced by Fourier integral. Hence the joint sine and cosine transform. Weber transform and Hankel transform and many numerous other transforms are obtained in this manner. The ordinary cosine transform $K(x,\lambda)=\sqrt{2/\pi}\cos(\lambda x)$ is just a special case. – Nemo Nov 25 '15 at 10:33