1. ## Difficult Integral

Hi I was wondering if anyone happen to know how to evaluate this integral. Thanks.

Question: Show that $\displaystyle\int_0^\infty\frac{e^{-\sqrt{2}t}\sinh t\sin t}{t}dt=\frac{\pi}{8}$

2. Originally Posted by nonsingular
Hi I was wondering if anyone happen to know how to evaluate this integral. Thanks.

Question: Show that $\displaystyle\int_0^\infty\frac{e^{-\sqrt{2}t}\sinh t\sin t}{t}dt=\frac{\pi}{8}$
Consider $I(m) = \int_0^{\infty} \frac{e^{-mt}\sinh t \sin t}{t}\,dt$ assuming m > 1. So the answer you require is $I(\sqrt{2})$. Calculating

$
I'(m) = - \int_0^{\infty} t\,\frac{e^{-mt}\sinh t \sin t}{t}\,dt = - \int_0^{\infty} e^{-mt}\sinh t \sin t\,dt
$

$
= - \frac{e^{-mt}}{m^4+4}
$
$
\left. \left(m^3 \sinh t \sin t + 2m \cosh t \cos t +(m^2+2) \sinh t \cos t + (m^2-2) \cosh t \sin t \right)\right|_0^{\infty}
$

$= \frac{2m}{m^4+4}$ provided that $m > 1$ which we assumed. Hence,

$
I(m) = \frac{1}{2} \tan^{-1}\frac{m^2}{2}
$
from which it follows that $I(\sqrt{2}) = \frac{1}{2} \tan^{-1} 1 = \frac{\pi}{8}.$

3. I would write $\sinh t={e^t-e^{-t}\over 2}$ and then use the tables for the integrals of $e^t$ times $\sin t$.

Or do it via parts if you wish.

Opps, I didn't see the t in the denominator, nevermind.

4. Originally Posted by Danny
Consider $I(m) = \int_0^{\infty} \frac{e^{-mt}\sinh t \sin t}{t}\,dt$ assuming m > 1. So the answer you require is $I(\sqrt{2})$. Calculating

$
I'(m) = - \int_0^{\infty} t\,\frac{e^{-mt}\sinh t \sin t}{t}\,dt = - \int_0^{\infty} e^{-mt}\sinh t \sin t\,dt
$

$
= - \frac{e^{-mt}}{m^4+4}
$
$
\left. \left(m^3 \sinh t \sin t + 2m \cosh t \cos t +(m^2+2) \sinh t \cos t + (m^2-2) \cosh t \sin t \right)\right|_0^{\infty}
$

$= \frac{2m}{m^4+4}$ provided that $m > 1$ which we assumed. Hence,

$
I(m) = \frac{1}{2} \tan^{-1}\frac{m^2}{2}
$
from which it follows that $I(\sqrt{2}) = \frac{1}{2} \tan^{-1} 1 = \frac{\pi}{8}.$
i didn't check what you got for $I'(m).$ assuming that it's correct, we can only say that $I(m)=\frac{1}{2} \tan^{-1} \frac{m^2}{2} + C,$ for some constant $C.$

here's my solution: let $J(a,b)=\int_0^{\infty}\frac{e^{-at}-e^{-bt}}{t}\sin t \ dt,$ where $a >0, \ b > 0.$ then:

$J(a,b)=\int_0^{\infty}\sin t \int_a^b e^{-xt} \ dx \ dt=\int_a^b \int_0^{\infty}e^{-xt} \sin t \ dt \ dx=\int_a^b \frac{1}{x^2+1} \ dx$

$=\tan^{-1}b - \tan^{-1}a=\tan^{-1} \left(\frac{b-a}{1+ab} \right).$

now we have: $I(m)=\frac{1}{2}J(m-1,m+1)=\frac{1}{2} \tan^{-1} \left(\frac{2}{m^2} \right).$ thus: $I(\sqrt{2})=\frac{1}{2} \tan^{-1}1 = \frac{\pi}{8}.$

5. Thanks for all the replies everyone. I was wondering if I can do this:
Let $\displaystyle I_1(s)=\int_0^\infty \frac{e^{-st}\sinh t\sin t}{t}dt$.
Then we let $\displaystyle I_2=\int_0^\infty e^{-st}\sinh t\sin t dt=\int_0^\infty e^{-st}\left(\frac{e^t-e^{-t}}{2}\right)\sin t dt$.
So $\displaystyle I_2=\frac{1}{2}\int_0^{\infty}e^{-t(s-1)}\sin tdt-\frac{1}{2}\int_0^\infty e^{-t(s+1)}\sin t dt=\frac{1}{2}\left [\frac{1}{(s-1)^2+1}-\frac{1}{(s+1)^2+1}\right ]$
Next, we use $\displaystyle\mathcal{L}\left\{\frac{f(t)}{t}\righ t\}=\int_s^{\infty}F(u)du$ where $F(s)=\mathcal{L}\{f(t)\}$
Hence $\displaystyle I_1(s)=\mathcal{L}\left\{\frac{\sinh t\sin t}{t}\right\} =\frac{1}{2}\int_s^\infty\frac{1}{(m-1)^2+1}dm-\frac{1}{2}\int_s^\infty\frac{1}{(m+1)^2+1}dm$
This simplifies to $\displaystyle I_1(s)=\frac{1}{2}[-\tan^{-1}(s-1)+\tan^{-1}(s+1)]$.
In our original problem, $s=\sqrt{2}$. Therefore, $\displaystyle I_1(\sqrt{2})=\frac{1}{2}\left[-\frac{\pi}{8}+\frac{3\pi}{8}\right ] =\frac{\pi}{8}$

6. Originally Posted by nonsingular
Thanks for all the replies everyone. I was wondering if I can do this:
Let $\displaystyle I_1(s)=\int_0^\infty \frac{e^{-st}\sinh t\sin t}{t}dt$.
Then we let $\displaystyle I_2=\int_0^\infty e^{-st}\sinh t\sin t dt=\int_0^\infty e^{-st}\left(\frac{e^t-e^{-t}}{2}\right)\sin t dt$.
So $\displaystyle I_2=\frac{1}{2}\int_0^{\infty}e^{-t(s-1)}\sin tdt-\frac{1}{2}\int_0^\infty e^{-t(s+1)}\sin t dt=\frac{1}{2}\left [\frac{1}{(s-1)^2+1}-\frac{1}{(s+1)^2+1}\right ]$
Next, we use $\displaystyle\mathcal{L}\left\{\frac{f(t)}{t}\righ t\}=\int_s^{\infty}F(u)du$ where $F(s)=\mathcal{L}\{f(t)\}$
Hence $\displaystyle I_1(s)=\mathcal{L}\left\{\frac{\sinh t\sin t}{t}\right\} =\frac{1}{2}\int_s^\infty\frac{1}{(m-1)^2+1}dm-\frac{1}{2}\int_s^\infty\frac{1}{(m+1)^2+1}dm$
This simplifies to $\displaystyle I_1(s)=\frac{1}{2}[-\tan^{-1}(s-1)+\tan^{-1}(s+1)]$.
In our original problem, $s=\sqrt{2}$. Therefore, $\displaystyle I_1(\sqrt{2})=\frac{1}{2}\left[-\frac{\pi}{8}+\frac{3\pi}{8}\right ] =\frac{\pi}{8}$
that's exactly my solution written in a slightly different way! so you can't be wrong! haha ... you could've simplified your result using the identity: $\tan^{-1}(s+1)-\tan^{-1}(s-1)=\tan^{-1}(2/s^2).$

regarding Danny's solution: i think $I'(m)=\frac{-2m}{m^4+4}$ and not $\frac{2m}{m^4 + 4}$ (hopefully somebody will check this!). then his solution can be fixed: $I(m)=\frac{-1}{2}\tan^{-1} \left(\frac{m^2}{2}\right) + C$ and $\lim_{m\to\infty}I(m)=0.$

thus $C=\frac{\pi}{4}$ and hence $I(m)=\frac{1}{2} \left[\frac{\pi}{2} - \tan^{-1}\left(\frac{m^2}{2}\right) \right]=\frac{1}{2}\tan^{-1}\left(\frac{2}{m^2}\right),$ which is exactly what i got in my previous post

7. Originally Posted by NonCommAlg
that's exactly my solution written in a slightly different way! so you can't be wrong! haha ... you could've simplified your result using the identity: $\tan^{-1}(s+1)-\tan^{-1}(s-1)=\tan^{-1}(2/s^2).$

regarding Danny's solution: i think $I'(m)=\frac{-2m}{m^4+4}$ and not $\frac{2m}{m^4 + 4}$ (hopefully somebody will check this!). then his solution can be fixed: $I(m)=\frac{-1}{2}\tan^{-1} \left(\frac{m^2}{2}\right) + C$ and $\lim_{m\to\infty}I(m)=0.$

thus $C=\frac{\pi}{4}$ and hence $I(m)=\frac{1}{2} \left[\frac{\pi}{2} - \tan^{-1}\left(\frac{m^2}{2}\right) \right]=\frac{1}{2}\tan^{-1}\left(\frac{2}{m^2}\right),$ which is exactly what i got in my previous post
I've checked and I missed the minus sign (and the constant OMG).

Thanks for cleaning things up.