Question

Integration help!

Answer 1

\[\int\limits_{1}^{\infty}dx/(x^5(e^{1/x}-1))\]

Answer 2

wut grade r u in

Answer 3

first year university

Answer 4

Doesn't look like an analytic solution. It can be solved by series, or numerical. Are you in any of these?

Answer 5

show that \[e^t \ge 1+t\] for \[t \ge0\] hence explain briefly why the integral must converge.

Answer 6

By MacLaurin's expansion,
\( e^t \) = 1+t/1!+t^2/2!+t^3/3! +...
=>
\( e^t \ge \) 1+t for t\( \gt 0 \)
This means that
\( e^t-1 \ge \) 1+t-1 =t for t\( \gt 0 \)
or
\( 1/(e^t-1) \le \) 1/t for t\( \gt 0 \)
therefore
I < integral of 1/x^6

Answer 7

and
\[\int\limits_{1}^{\infty} \frac{dx}{x^6} \ = \ \frac{1}{5}\]

Answer 8

i dont think we can use MacLaurin's expansion though

Answer 9

Then you can show that e^t=1 when t=0.
and d(e^t)/dt >1 for t>0, therefore
e^t-1 >0 for t>0.

Answer 10

ok thanks