1. ## differential equations

If a and b are positive constants, evaluate the following integrals:

The integral from 0 to infinity, of (e^-ax sin bx)/x dx

ta said answer was: pie/2 - tan^-1 (a/b), but book says tan^-1(b/a), so which? and how?

The integral from 0 to infinity, of (e^-ax - e^-bx)/x dx

ta said answer was ln(a/b), but book says log b/a, so which? and how?

thanks!

2. Originally Posted by dixie
If a and b are positive constants, evaluate the following integrals:

The integral from 0 to infinity, of (e^-ax sin bx)/x dx

ta said answer was: pie/2 - tan^-1 (a/b), but book says tan^-1(b/a), so which? and how?

The integral from 0 to infinity, of (e^-ax - e^-bx)/x dx

ta said answer was ln(a/b), but book says log b/a, so which? and how?

thanks!
first, what does this have to do with differential equations? secondly, are you sure the /x should be there? example, $\int e^{-ax} \sin bx ~dx$ can be worked out, but $\int \frac {e^{-ax} \sin bx}x ~dx$ has no solution in elementary functions. it's an elliptical integral i believe

also, be careful with what your TA says and make sure all the signs are right. for instance, -ln(a/b) = ln(b/a). so make sure you're not leaving off a minus sign or something

3. Originally Posted by dixie
If a and b are positive constants, evaluate the following integrals:

The integral from 0 to infinity, of (e^-ax sin bx)/x dx
First, $\int e^{cx}\sin bx dx = \frac{e^{cx}}{c^2+b^2} \left( c\sin bx - b\cos x \right) + k$ for $b,c\not = 0$.

Therefore, if $c<0$ then $\int \limits_0^{\infty} e^{cx} \sin bx dx = \frac{e^{cx}}{c^2+b^2} \left( c\sin bx - b\cos x \right) \bigg|_0^{\infty} = \frac{b}{b^2+c^2}$

Also, $\int_0^{\infty} e^{-xy} dy = \frac{1}{x}$ for $x>0$.

Now, $\int_0^{\infty} \frac{e^{-ax}\sin bx}{x} dx = \int \limits_0^{\infty} \int \limits_0^{\infty} e^{-ax}\sin (bx) e^{-xy} dy ~ dx$

Change integration order, $\int \limits_0^{\infty} \int \limits_0^{\infty} e^{-x(a+y)}\sin (bx) dx ~ dy = \int \limits_0^{\infty} \frac{b}{b^2 + (a+y)^2}dy$

But, $\int \limits_0^{\infty} \frac{b}{b^2 + (a+y)^2}dy = \int \limits_a^{\infty} \frac{b}{b^2 + t^2} dt = \tan^{-1}\frac{t}{b} \bigg|_a^{\infty} = \frac{\pi}{2} - \tan^{-1} \frac{a}{b}$

Which is the same as $\tan^{-1} \frac{b}{a}$ because $\tan^{-1} \tfrac{1}{x} = \tfrac{\pi}{2} - \tan x,x>0$

4. Originally Posted by dixie
The integral from 0 to infinity, of (e^-ax - e^-bx)/x dx
Again, $\frac{1}{x} = \int_0^{\infty} e^{-yx} dy$ for $x>0$.

Thus, $\int_0^{\infty} \frac{e^{-ax} - e^{-bx}}{x} dx = \int \limits_0^{\infty} \int \limits_0^{\infty} \left( e^{-ax} - e^{-bx} \right) e^{-yx} dy ~ dx$.

Switch order, $\int \limits_0^{\infty}\int \limits_0^{\infty} e^{-(a+y)x} - e^{-(b+y)x} dx ~ dy = \int \limits_0^{\infty} \frac{1}{a+y} - \frac{1}{b+y}dy = \log \left( \frac{a+y}{b+y} \right) \bigg|_0^{\infty}$

Thus, we get, $\log 1 - \log \tfrac{a}{b} = - \log \tfrac{a}{b} = \log \tfrac{b}{a}$

5. It's actually quickier by considering $\frac{{{e}^{-ax}}-{{e}^{-bx}}}{x}=\int_{a}^{b}{{{e}^{-xy}}\,dy}.$

And of course, the interchange of order of integration is justified by Tonelli's Theorem for non-negative functions.