Question

integrate between limits 0 and d50〖1/(√2π *dp* ln GSD) exp[-〖{ln(dp )-ln(MMD)}〗^2/(2〖{ln(GSD)}〗^2 )]ddp 〗

Answer 1

$\int_0^\infty\frac{1}{\sqrt{2π}ln(a)}\frac{e^{-\frac{{ln(\frac xb)}^2}{2{ln(b)}^2}}}{x}dx,$ $x=dp;a=GSD;b=MMD$

Substitute $ln(\frac{x}{b})=y;\frac{dx}{x}=dy$

$\frac{1}{\sqrt{2π}ln(a)}\int_{-\infty}^\infty e^{-\frac{{y}^2}{2{ln(b)}^2}}dy$

Substitute $\frac{y}{\sqrt{2}ln(b)}=z;dz=\frac{dy}{\sqrt2ln(b)}$

$\frac{1}{\sqrt{2π}ln(a)}\int_{-\infty}^\infty e^{-z^2}(\sqrt2ln(b))dz$

$2\frac{log _ab}{\sqrt{π}}\int_0^\infty e^{-z^2}dz$

Substitute $z^2=t;2zdz=dt$

$\frac{log _ab}{\sqrt{π}}\int_0^\infty e^{-t}t^{-\frac{1}{2}}dt$

$\frac{log _ab}{\sqrt{π}}\Gamma(\frac{1}{2})=\frac{log _ab}{\sqrt{π}}\sqrt{π}= {log _ab}(Answer)$