Finding Integral Value: \int_{1}^{e^\frac{\pi}{2}} \sin (lnx) dx

[Doc G]

Hi,

Homework Statement

find value of
$$\int_{1}^{e^\frac{\pi}{2}} \sin (lnx) dx$$

Homework Equations

The Attempt at a Solution

I first used subsitution u = lnx, that gave me:

$$\int_{0}^{\frac{\pi}{2}} e^{u}\sin u du$$
then letting
$$I = \int_{0}^{\frac{\pi}{2}} e^{u}\sin u du$$
integrating by parts twice, gave me:
$$I = - e^{u} \cos u +\int_{0}^{\frac{\pi}{2}} u e^{u}\cos u du$$
$$I = - e^{u} \cos u + e^{u} \cos u + \int_{0}^{\frac{\pi}{2}} e^{u}\sin u du$$

therefore

$$I = - e^{u} \cos u + e^{u} \cos u + I$$
which of course cancels to zero which is not very useful

Any ideas on finding the value of this integral, or have i made an error in my working?

Many Thanks

 

[neutrino]

I think you need to re-check the 'by-parts'.

 

[CompuChip]

Yep. In particular, in the second "by parts", think where that sine inside the integral comes from :)

 

[Doc G]

Taking the result of the first integration by parts:

$$I = - e^{u} \cos u +\int_{0}^{\frac{\pi}{2}} u e^{u}\cos u du$$

then working with the integral part:
$$+\int_{0}^{\frac{\pi}{2}} u e^{u}\cos u du$$
let u= cos u
therefore
$$\frac{du}{dx}= - sinu$$
let
$$\frac{dv}{dx}= u e^{u}$$
therefore
$$v = e^{u}$$

then using identity that:

$$\int u \frac{dv}{dx} \equiv uv - \int v \frac{du}{dx}$$

I got
$$e^{u} \cos u + \int_{0}^{\frac{\pi}{2}} e^{u}\sin u du$$

Note- i chose u=cos u and dv/dx =ue^u [rather than u=ue^u]since i thought otherwise I would never be able to achieve an equivalent expression without an integral involved as I would get greater and greater powers of u in front of e^u as I kept going.

Sorry if I'm not seeing something obvious, but i can't find the error there

thanks for the replies

 

[neutrino]

Note- i chose u=cos u and dv/dx =ue^u [rather than u=ue^u]since i thought otherwise, I thought I would never be able to achieve an equivalent expression without an integral involved as I would get great and great powers of u in front of e^u as I kept going.

How did you get a 'u' within the integral at the result of the first integration by parts? (And it's now the derivative with respect to u, not x)

[I'm using t instead of u for the sake of clarity]

The first integral, after substitution $$= \int_{0}^{\pi/2}{e^t\sin{t}}dt$$

Let u = cos(t) and dv/dt = e^t. Therefore, du/dt = -sin(t) and v = e^t.

 

[Doc G]

Thanks, I think I've got it now; but just to check here my full working:

let
$$I = \int_{0}^{\frac{\pi}{2}} e^{t}\sin t dt$$

u = sin t
dv/dt= e^t

therefore
du/dt = cos t
v = e^t

so:
$$I = e^{t} sin t - \int_{0}^{\frac{\pi}{2}} e^{t} cos t dt$$

and integrating by parts again:
u = cos t
dv/dt = e^t

so
du/dt = -sin t
v = e^t

$$I = e^{t} sin t - e^{t} cos t - \int_{0}^{\frac{\pi}{2}} e^{t} sin t dt$$

$$I = e^{t} sin t - e^{t} cos t - I$$

$$2I = [e^{t} ( sin t - cos t )]_{0}^{\frac{\pi}{2}}$$
$$I = \frac{1}{2} [(e^{\frac{\pi}{2}}( 1 - 0 )) - ( -1)]$$

and finally, giving
$$I = \frac{1}{2} [e^{\frac{\pi}{2}} + 1 ]$$

Hopefully its right this time?

Thanks again

 

[neutrino]

Perfect! ...

 

[Doc G]

Great! Thank you very much for the advice

 

[neutrino]

You're welcome. :)

 

[CompuChip]

Hmm, I don't quite follow what you're doing in your subsequent posts (it looks correct though), but your first try was already correct if it weren't for one minus sign:

$$\int_{1}^{e^\frac{\pi}{2}} \sin (lnx) dx$$

I first used subsitution u = lnx, that gave me:

$$\int_{0}^{\frac{\pi}{2}} e^{u}\sin u du$$
then letting
$$I = \int_{0}^{\frac{\pi}{2}} e^{u}\sin u du$$
integrating by parts twice, gave me:
$$I = - e^{u} \cos u +\int_{0}^{\frac{\pi}{2}} u e^{u}\cos u du$$
$$I = - e^{u} \cos u + e^{u} \cos u - \int_{0}^{\frac{\pi}{2}} e^{u}\sin u du$$
Note the minus sign: cos' = -sin

therefore

$$I = - e^{u} \cos u + e^{u} \cos u - I$$

so
$$2I = - e^{u} \cos u + e^{u} \cos u$$
and the answer follows.

 

[neutrino]

Hmm, I don't quite follow what you're doing in your subsequent posts (it looks correct though), but your first try was already correct if it weren't for one minus sign:


so
$$2I = - e^{u} \cos u + e^{u} \cos u$$
and the answer follows.

And you end up with 0, which is incorrect.