Question

How do I integrate this
∫e^(x-1)/x^2
What is this wolfram alpha result? Am I doing real math now?
Also, how would I evaluate that result at 1?

Answer 1

use an integration table would be my bet

Answer 2

or maybe integration by parts

Answer 3

Not a polynomial power , wouldn't eliminate it.

Answer 4

\(\color{#000000}{ \displaystyle \int \frac{e^{x-1} }{x^2} dx }\)

my first bet is a power series.

Answer 5

i was thinking we might get a back substitution since the cube would cancel. Maybe

Answer 6

\(\color{#000000}{ \displaystyle e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!} }\)

\(\color{#000000}{ \displaystyle e^{x-1}=\frac{1}{e}\sum_{k=0}^{\infty}\frac{x^k}{k!} }\)

\(\color{#000000}{ \displaystyle x^{-2}e^{x-1}=\frac{1}{e}\sum_{k=0}^{\infty}\frac{x^{k-2}}{k!} }\)

\(\color{#000000}{ \displaystyle \int x^{-2}e^{x-1}~dx=\frac{1}{e}\int \sum_{k=0}^{\infty}\frac{x^{k-2}}{k!} dx }\)

\(\color{#000000}{ \displaystyle \int x^{-2}e^{x-1}~dx=\frac{1}{e}\sum_{k=0}^{\infty}\frac{x^{k-1}}{(k-1)k!} +C }\)

Then, there is a domain problem. Always annoyed me with power series.

Answer 7

I don't even remember those O.O You go Zelman!

Answer 8

\(\color{#000000}{ \displaystyle \int x^{-2}e^{x}~dx=\ }\)

\(\color{#000000}{ \displaystyle e^x=\sqrt{u} \quad\Longleftarrow \quad x=\sqrt{\ln(u)}\quad\Longrightarrow \quad x^2=\ln (u) }\)

\(\color{#000000}{ \displaystyle dx=\frac{1}{2u\sqrt{\ln(u)}}~du }\)

\(\color{#000000}{ \displaystyle \frac{1}{2} \int\limits_{}^{}\frac{\sqrt{~\ln u}}{\sqrt{~u}}~du }\)

I attempted a wild substitution, but I suppose that power-series integration is the only approach here.

\(\color{#000000}{ \displaystyle \int x^{-2}e^{x}~dx=\int x^{-2}(\cos x - i\sin x)~dx= }\)

the sine integral for the x\(^{-2}\)cos(x), and cosine integral forn the \(\small i\)x\(^{-2}\)sin(x).

Answer 9

Oh, sorry, you add the isin(x).