I solved $$\int_0^\infty \frac{e^{-b/x} }{x^{a+1}}.x.\frac{b^{a-1}}{\Gamma(a-1)}dx$$ and got result -1 while https://www.integral-calculator.com/ showed result 1. Can someone confirm which one the correct result?
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So many confusing things. What is $.$ there? Is it multiplication? Why to you keep $\frac{b^{a-1}}{\Gamma(a-1)}$ inside the integral? (It's a constant and it doesn't depend on $x$.) // Anyway if $.$ is multiplication, one can rewrite the integral as: $$\frac{b^{a-1}}{\Gamma(a-1)}\int_0^1 \frac{e^{-b/x}}{x^{a}}dx$$ Now let $x=\frac{1}{t}$ then use the gamma function integral representation. Let me know if you need further help. – Zacky Jun 21 '19 at 09:12
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Yes, that's just multiplication and I have got the final result (=-1). The question is; "Does my result correct?" It's integral from 0 to infinite, not 0 to 1! – Alexander Kevin Gilbert Jun 21 '19 at 09:38
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Yes, it is correct. I typed $1$ instead of $\infty$ in a hurry. – Zacky Jun 21 '19 at 09:38