How would you integrate ∫x^(-1/2)e^(-bx)?

Question

I know that I should do integration by parts for problems similar to this one, for instance if the first term was raised to a whole number power, in which case that term would eventually simplify, but this case doesn't seem to...

Answer 1

Substituting u²=x leads to ∫e^(-bu²)du which has no solution as an indefinite integral. But a solution is available as a series:

∫(e^-(bx)/√x)dx=∫(1/√x)(1-bx+b²x²/2!-b³x³/3!-...(-1)ⁿbⁿxⁿ/n!+...)dx=

C+ʃ(1/√x-b√x+b²x^(3/2)/2!-b³x^(5/2)/3!-...(-1)ⁿbⁿx^(n-½)/n!+...)dx=

C+2∑(-1)ⁿbⁿx^(n+½)/(n!(2n+1)) for integer n≥0 and C is the integration constant.