"\\int_0^\\infty\\frac{1}{\\sqrt{2\u03c0}ln(a)}\\frac{e^{-\\frac{{ln(\\frac xb)}^2}{2{ln(b)}^2}}}{x}dx," "x=dp;a=GSD;b=MMD"
Substitute "ln(\\frac{x}{b})=y;\\frac{dx}{x}=dy"
"\\frac{1}{\\sqrt{2\u03c0}ln(a)}\\int_{-\\infty}^\\infty e^{-\\frac{{y}^2}{2{ln(b)}^2}}dy"
Substitute "\\frac{y}{\\sqrt{2}ln(b)}=z;dz=\\frac{dy}{\\sqrt2ln(b)}"
"\\frac{1}{\\sqrt{2\u03c0}ln(a)}\\int_{-\\infty}^\\infty e^{-z^2}(\\sqrt2ln(b))dz"
"2\\frac{log _ab}{\\sqrt{\u03c0}}\\int_0^\\infty e^{-z^2}dz"
Substitute "z^2=t;2zdz=dt"
"\\frac{log _ab}{\\sqrt{\u03c0}}\\int_0^\\infty e^{-t}t^{-\\frac{1}{2}}dt"
"\\frac{log _ab}{\\sqrt{\u03c0}}\\Gamma(\\frac{1}{2})=\\frac{log _ab}{\\sqrt{\u03c0}}\\sqrt{\u03c0}=\n{log _ab}(Answer)"
Comments
Dear azike. Thank you for a leaving a feedback. It was not clear to which exactly expression the square root sign should be applied, we considered one possible case. If you need another case, then submit a new question and describe some details in words to avoid a confusion. If the upper and lower limits will be interchanged, then the new definite integral is equal to the previous definite integral taken with the minus sign.
i am so grateful that you people were able to find solution to my mathematical problem, thank you very much. there is a little problem with the solution. the first line is not clear. the exponential why all over x. x is dp and is supposed to be with square root of 2pi*x*ln(a). also upper limit is zero(0) while lower limit is d50. thanks.
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