Definite Integral of modified bessel function of second kind

Question

How do I integrate $$\int_{0}^{d} \sqrt{\frac{a-bx}{c}}K_1\left(\sqrt{\frac{a-bx}{c}}\right) dx$$ where $K_1$ represents modified Bessel function of second kind and $a,b,c,d$ are constants?

Please help.

Answer 1

$$\int_{0}^{d}\!\sqrt {{\frac {-bx+a}{c}}}{{K}_{1}\left(\sqrt {{ \frac {-bx+a}{c}}}\right)}\,{\rm d}x $$ substituting: ${\frac {-bx+a}{c}}=t$

$$-{\frac {c}{b}\int_{{\frac {a}{c}}}^{{\frac {-bd+a}{c}}}\!\sqrt {t}{ {K}_{1}\left(\sqrt {t}\right)}\,{\rm d}t} $$ substituting: $t={x}^{2}$

$$-{\frac {c}{b}\int_{\sqrt {{\frac {a}{c}}}}^{\sqrt {{\frac {-bd+a}{c}} }}\!2\,{x}^{2}{{K}_{1}\left(x\right)}\,{\rm d}x} $$

By parts: $$-2\,{{K}_{0}\left(\sqrt {-{\frac {bd}{c}}+{\frac {a}{c}}}\right)}d +2\,{\frac {a}{b}{{K}_{0}\left(\sqrt {-{\frac {bd}{c}}+{\frac {a}{ c}}}\right)}}-2\,{\frac {a}{b}{{K}_{0}\left(\sqrt {{\frac {a}{c}}} \right)}}-4\,{\frac {c}{b}\int_{\sqrt {{\frac {a}{c}}}}^{\sqrt {{ \frac {-bd+a}{c}}}}\!{{K}_{0}\left(x\right)}x\,{\rm d}x}$$

and by parts: $$\int_{\sqrt {{\frac {a}{c}}}}^{\sqrt {{\frac {-bd+a}{c}}}}\!{{K}_{0 }\left(x\right)}x\,{\rm d}x={{K}_{1}\left(\sqrt {{\frac {a}{c}}} \right)}\sqrt {{\frac {a}{c}}}-{{K}_{1}\left(\sqrt {{\frac {-bd+a }{c}}}\right)}\sqrt {{\frac {-bd+a}{c}}}$$

then we have:

$\color{Blue}{\int_{0}^{d}\!\sqrt {{\frac {-bx+a}{c}}}{{K}_{1}\left(\sqrt {{ \frac {-bx+a}{c}}}\right)}\,{\rm d}x=\\{\frac {1}{b} \left( \left( -2\,bd+2\,a \right) {{K}_{0}\left( \sqrt {{\frac {-bd+a}{c}}}\right)}+4\,{{K}_{1}\left(\sqrt {{\frac {-bd+a}{c}}}\right)}\sqrt {{\frac {-bd+a}{c}}}c-4\,{{K}_{1}\left( \sqrt {{\frac {a}{c}}}\right)}\sqrt {{\frac {a}{c}}}c-2\,{{K}_{0} \left(\sqrt {{\frac {a}{c}}}\right)}a \right) }}$

Maple code:

int(sqrt((-b*x + a)/c)*BesselK(1, sqrt((-b*x + a)/c)), x = 0 .. d) = ((-2*b*d + 2*a)*BesselK(0, sqrt((-b*d + a)/c)) + 4*BesselK(1, sqrt((-b*d + a)/c))*sqrt((-b*d + a)/c)*c - 4*BesselK(1, sqrt(a/c))*sqrt(a/c)*c - 2*BesselK(0, sqrt(a/c))*a)/b