Question

Please help with this integral : e^(-x)/x^3 as x goes to infinity ... now i know the answer is infinity but i don't know how to solve it to get to the answer =\

Answer 1

integral or limit??

Answer 2

integral =)

Answer 3

integral from 0 to infinity or -infinity to infinity?

Answer 4

0 to infinity

Answer 5

have u tried uv(product) rule by making the function e^(-x)x^(-3)?

Answer 6

\[\int_{0}^{\infty} \frac{e^{-x}}{x^3} \text{d}x\]?

Answer 7

yes !

Answer 8

i think integration by parts will work here...as hartnn said

are u familiar with integration by parts?

Answer 9

yes i am .. but i got integral of e^(-x)*lnx eventually which i couldn't solve =\

Answer 10

howahhh...i found something finally...with integration by parts u will get :

\[\int_{0}^{\infty} \frac{e^{-x}}{x^3} \text{d}x=-\frac{e^{-x}}{2x^2} \mid_0^\infty+\frac{e^{-x}}{2x} \mid_0^\infty+\int_{0}^{\infty} \frac{e^{-x}}{x} \text{d}x\]check this then i will continue :)

Answer 11

lol... @PeX are u there?

Answer 12

there is a trick to do that

Answer 13

i think u should continue by taking 1/x term as the derivative unitill it reaches the power of 1/x3 then bring this integral to the left side and add with e^(-x)/x^3

Answer 14

nope it cancels everything...and we will have the same thing

Answer 15

\[\int_{0}^{\infty} \frac{e^{-x}}{x^3} \text{d}x=-\frac{e^{-x}}{2x^2} \mid_0^\infty+\frac{e^{-x}}{2x} \mid_0^\infty+\int_{0}^{\infty} \frac{e^{-x}}{x} \text{d}x\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\lim_{x \rightarrow 0} (\frac{1}{2x^2}-\frac{1}{2x})+\int_{0}^{\infty} \frac{e^{-x}}{x} \text{d}x=+\infty+\text{Something >0}=+\infty \]

Answer 16

\[\int\limits_{}^{}e^-x/x=?\]

now let 1/x=lnt
e^1/x=t
we have 1/x=lnt and e^-x=t substitute
\[\int\limits_{}^{}tlnt\]

= \[(t^2/2)lnt -\int\limits_{}^{}t/2\]
\[t^2lnt/2-t^/4\]

Answer 17

there should be -t^2/4 last expression

Answer 18

sorry i tried it's extremely wrong

Answer 19

Thank you very much !!! you are the math messiah =)

Answer 20

yw...:)