Integrating a natural log in double integral

[Emspak]

Homework Statement

$\int^{1}_{0}$$\int$$^{e^x}_{e^-x}$$\frac{lny}{y}$dydx

The attempt at a solution

So I am integrating ln(y)/y and I tried it by parts, first with u = ln(y), dv = 1/y, and therefore du = 1/y, and v = ln y

but if I use that I get

(ln(y))²-$\int$$\frac{lny}{y}$ again.

So I tried switching u and v around. I got: u = 1/y, dv = ln(y) dy, du = 1/y² and v = 1/y.

On that basis I get $\frac{1}{y^2}$-$\int$$\frac{1}{y^3}$|$^{e^x}_{e^-x}$dx

and from there I get

$\int^{1}_{0}$ $\frac{1}{y^2}$-$\frac{1}{4y^4}$|$^{e^x}_{e^-x}$dx

But I have a sneaking suspicion I have done something horribly wrong. I see the integral I want as (ln(y))²/2. But that doesn't make sense to me.

So, I did something messed up. If someone could tell me where it is, that would be much appreciated.

 

[SammyS]

Homework Statement

$\int^{1}_{0}$$\int$$^{e^x}_{e^-x}$$\frac{lny}{y}$dydx

The attempt at a solution

So I am integrating ln(y)/y and I tried it by parts, first with u = ln(y), dv = 1/y, and therefore du = 1/y, and v = ln y

Try substitution.

Let u = ln(y).

but if I use that I get

(ln(y))²-$\int$$\frac{lny}{y}$ again.

So I tried switching u and v around. I got: u = 1/y, dv = ln(y) dy, du = 1/y² and v = 1/y.

On that basis I get $\frac{1}{y^2}$-$\int$$\frac{1}{y^3}$|$^{e^x}_{e^-x}$dx

and from there I get

$\int^{1}_{0}$ $\frac{1}{y^2}$-$\frac{1}{4y^4}$|$^{e^x}_{e^-x}$dx

But I have a sneaking suspicion I have done something horribly wrong. I see the integral I want as (ln(y))²/2. But that doesn't make sense to me.

So, I did something messed up. If someone could tell me where it is, that would be much appreciated.

 

[Emspak]

I did that the first time, and showed what I got. u=ln y so du = 1/y, correct? Or not? If that's so and I et dv = 1/y then v has to be ln y, no? "try substitution" is what I just did.

 

[SammyS]

I did that the first time, and showed what I got. u=ln y so du = 1/y, correct? Or not? If that's so and I et dv = 1/y then v has to be ln y, no?

No.

You tried integration by parts.

Do integration by substitution, with the substitution u = ln(y). No v is needed.

 

[Emspak]

ah, i see (I was thinking u-substitutions in integration by parts) -- NOW It is a lot clearer. Thanks much.