How can I find the first term of the series expansion of
∫∞0ln(t)√te−ntdt,n→∞?
Or:
As ∫∞0ln(t)√te−ntdt=1√n∫∞0ln(tn)√te−tdt
What is lim
How to find \lim_{h\rightarrow 0}\frac{\sin(ha)}{h} without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
No comments:
Post a Comment