I need to show that
$\int_{0}^ady \int_{y}^a \sqrt{x^2-y^2}f'''(a-x)dx=\frac{\pi}2(f(a)-f(0)-af'(0)-\frac{a^2}2f''(0))$
I started first by doing integration by parts
$\int_{0}^ady \int_{y}^a \sqrt{x^2-y^2}f'''(a-x)dx=\int_{0}^a[-f''(a-x)\sqrt{x^2-y^2}]_{y}^ady+\int_{0}^ady \int_{y}^a f''(a-x)\frac{x}{\sqrt{x^2-y^2}}dx$
$=-f''(0)\frac{\pi a^2}4+\int_{0}^ady \int_{y}^a f''(a-x)\frac{x}{\sqrt{x^2-y^2}}dx$
I thought I was on the right track, but then when I do integration by parts once again
$\int_{0}^ady \int_{y}^a f''(a-x)\frac{x}{\sqrt{x^2-y^2}}dx=\int_{0}^a[-f'(a-x)\frac{x}{\sqrt{x^2-y^2}}]_{y}^a dy+\int_{0}^ady \int_{y}^a f'(a-x)\frac{y^2}{(x^2-y^2)^{3/2}}dx$
When I plug in the $y$, I get a diverging result
Am I missing something? or is there another way to show this besides the integration by parts?
Any help would very much be appreciated
Switching the order of integration we get
$$\int_0^af'''(a-x)dx\int_0^x\sqrt{x^2-y^2}dy = \frac{\pi}{4}\int_0^ax^2f'''(a-x)dx$$
since it's just the area of a quarter circle with radius $x$. From here integration by parts should get the desired equality.