Thread: Show that sech integral is equal to pi/2

1. Show that sech integral is equal to pi/2

I'm wondering if anyone has any advice on where to even start for proving this:

$\displaystyle \int_0^{\infty} sech(x) dx = \frac{\pi}{2}$

2. It is asked to prove that $\displaystyle \int_0^\infty {\frac{2} {{e^x + e^{ - x} }}\,dx} = \frac{\pi } {2}.$

And this does not take much time, 'cause it's a hidden arctangent, so

$\displaystyle \int_0^\infty {\frac{2} {{e^x + e^{ - x} }}\,dx} = 2\int_0^\infty {\frac{{e^x }} {{e^{2x} + 1}}\,dx} = 2 \cdot \arctan e^x \Big|_0^\infty = \frac{\pi } {2}.$

3. Thanks a lot! I'll try something similar to that for sech^2 and see what I get.

But why isn't that -pi/2? I mean isn't $\displaystyle \arctan e^x \big|^{\infty}_0 = - \frac{\pi}{4}$

4. No, be careful:

$\displaystyle \arctan \infty - \arctan 1 = \frac{\pi } {2} - \frac{\pi } {4} = \frac{\pi } {4}.$

5. Ok so my second question was:

$\displaystyle \int_0^{\infty} sech^2 (x) dx$

So I did:

$\displaystyle \int_0^{\infty} sech^2 (x) dx = tanh(x) \Big|^{\infty}_0 = \frac{sech(x)}{cosh(x)} \Big|^{\infty}_0$

$\displaystyle = \frac{e^x - e^{-x}}{e^x + e^{-x}} \Big|^{\infty}_0 = \frac{O(e^x)}{O(e^x)} - \frac{1 - 1}{1 + 1} = 1 - \tfrac{0}{2} = 1$

Does that look good?

6. Ok I've hit a brick wall with this one:

$\displaystyle I_n = \int_0^{\infty} sech^n (x) dx$

Show that:

$\displaystyle I_n = \frac{n-2}{n-1} I_{n-2}$

I'd just like a hint in the right direction.

7. My first thought is write $\displaystyle sech^n x$ as $\displaystyle sech^2 x sech^{n-2}x$

8. Ok so I tried that and it seems like it is somewhat on the right track but It doesn't divide by the (n-1):

$\displaystyle I_n = \int_0^{\infty} \underbrace{sech^2 (x)}_{\tfrac{d}{dx}tanh (x)} sech^{n-2} (x) dx$

$\displaystyle = tanh sech^{n-2} \Big|_0^{\infty} + \int_0^{\infty} tanh (x) (n-2) sech^{n-3} (x) sech (x) tanh (x) dx$

$\displaystyle = 0 + (n-2) \int_0^{\infty} tanh^2 (x) sech^{n-2} (x) dx$

$\displaystyle = (n-2) \int_0^{\infty} tanh^2 (x) sech^{n-2} (x) dx$

$\displaystyle = (n-2) \underbrace{\int_0^{\infty} sech^{n-2} (x) dx}_{I_{n-2}} + \int_0^{\infty} sech^n (x) dx$

9. Anyone have any ideas?

10. Originally Posted by TrevorP
Ok so I tried that and it seems like it is somewhat on the right track but It doesn't divide by the (n-1):

$\displaystyle I_n = \int_0^{\infty} \underbrace{sech^2 (x)}_{\tfrac{d}{dx}tanh (x)} sech^{n-2} (x) dx$

$\displaystyle = tanh sech^{n-2} \Big|_0^{\infty} + \int_0^{\infty} tanh (x) (n-2) sech^{n-3} (x) sech (x) tanh (x) dx$

$\displaystyle = 0 + (n-2) \int_0^{\infty} tanh^2 (x) sech^{n-2} (x) dx$

$\displaystyle = (n-2) \int_0^{\infty} tanh^2 (x) sech^{n-2} (x) dx$

$\displaystyle = (n-2) \underbrace{\int_0^{\infty} sech^{n-2} (x) dx}_{I_{n-2}} + \int_0^{\infty} sech^n (x) dx$
$\displaystyle I_n = \int_{0}^{\infty} sech^2 x \, \, sech^{n-2} x \, dx$.

Integration by parts is the way to go: $\displaystyle \int u \, dv = uv - \int v \, du$.

Let $\displaystyle u = sech^{n-2} \Rightarrow \frac{du}{dx} = sech^{n-3} x \times (-sech x \, \, tanh x)$

$\displaystyle \Rightarrow du = sech^{n-3} x \times (-sech x \, \, tanh x) \, dx = -sech^{n-2} x \, \, tanh x \, dx$.

Let $\displaystyle dv = sech^2 x \, dx \Rightarrow v = tanh x$.

Then $\displaystyle I_n = \left[ sech^{n-2} x \, \, tanh x\right]_0^\infty + (n-2) \int_0^\infty sech^{n-2} x \, \, tanh^2 x \, dx$.

Note that $\displaystyle sech^{n-2} x \, \, tanh^2 x = sech^{n-2} x \, \, (1 - sech^2 x) = sech^{n-2} x - sech^n x$.

Then $\displaystyle I_n = \left[ sech^{n-2} x \, \, tanh x\right]_0^\infty + (n-2) \int_0^\infty sech^{n-2} x - sech^n x \, dx$.

Therefore $\displaystyle I_n = \left[ sech^{n-2} x \, \, tanh x\right]_0^\infty + (n-2) (I_{n-2} - I_n)$.

Calculate $\displaystyle \left[ sech^{n-2} x \, \, tanh x\right]_0^\infty$ and sub it in.

Now re-arrange the resulting equation to solve for $\displaystyle I_n$ ......

11. Thanks a lot! I had actually tried tanh^2 thing but made a mistake. Here is the solution for anyone if they are interested (I just really like using this LaTeX stuff!):

$\displaystyle I_n = \int_0^{\infty} \underbrace{sech^2 (x)}_{\tfrac{d}{dx}tanh (x)} sech^{n-2} (x) dx$

$\displaystyle = tanh sech^{n-2} \Big|_0^{\infty} + \int_0^{\infty} tanh (x) (n-2) sech^{n-3} (x) sech (x) tanh (x) dx$

$\displaystyle = 0 + (n-2) \int_0^{\infty} tanh^2 (x) sech^{n-2} (x) dx$

$\displaystyle = (n-2) \int_0^{\infty} tanh^2 (x) sech^{n-2} (x) dx = (n-2) \int_0^{\infty} (1 - sech^2 (x)) sech^{n-2} (x) dx$

$\displaystyle = (n-2) \int_0^{\infty} \underbrace{sech^{n-2} (x)}_{I_{n-2}} - \underbrace{sech^n (x)}_{I_n} dx = (n-2)(I_{n-2} - I_n)$

$\displaystyle I_n + I_n(n-2) = (n-2)I_{n-2}$
$\displaystyle I_n (1 + n-2) = (n-2)I_{n-2}$
$\displaystyle I_n = \frac{(n-2)}{(n-1)}I_{n-2}$