I sloved the following problem today. I liked it, but it might be unfair because I used an integral which I already knew from some time ago.
$\displaystyle \int_0^1 \frac{\tan^{-1} x}{x+1} dx$
We were talkin' about this problem yesterday. I already knew it, and it can be tackled with double integration, but this is not so helpful, even doing integration by parts we'll get the same integral as by putting a parameter to construct a double integral and reverse integration order. I'll solve $\displaystyle \lambda$ with double integration.
$\displaystyle \int_0^1 {\frac{{\arctan x}}
{{x + 1}}\,dx} = \arctan x\cdot\ln(1+x)\bigg|_0^1 - \underbrace {\int_0^1 {\frac{{\ln (1 + x)}}
{{1 + x^2 }}\,dx} }_\lambda .$
We have $\displaystyle \lambda = \int_0^1 {\int_0^x {\frac{{dz}}
{{1 + z}}} \cdot \,\frac{{dx}}
{{1 + x^2 }}} .$ Now we'll try to get an unit square integral, so let $\displaystyle z=xy,$ then after reversing integration order $\displaystyle \lambda$ becomes $\displaystyle \int_0^1 {\int_0^1 {\frac{x}
{{1 + xy}} \cdot \frac{1}
{{1 + x^2 }}\,dx} \,dy}.$ Now, note that
$\displaystyle \begin{aligned}\frac{x}{{1+xy}}\cdot\frac{1}{{1+x^ 2}}&=\frac{1}{{1 + y^2 }}\cdot\frac{{x+xy^2}}{{(1+xy)\left({1+x^2}\right) }}\\&= \frac{1}{{1 + y^2 }}\Bigg\{ {\frac{{\left( {x^2y+x+xy^2+y} \right) - y\left({1 + x^2 } \right)}}{{(1 + xy)\left({1+x^2 }\right)}}}\Bigg\}\\&= \frac{1}{{1 + y^2 }}\Bigg\{ {\frac{{(1 +xy)(x+y)-y\left({1+x^2}\right)}}{{(1+xy)\left({1+x^2}\right )}}}\Bigg\},
\end{aligned}$
hence the inner integral is
$\displaystyle \int_0^1 {\left\{ {\frac{x}
{{1 + x^2 }} + \frac{y}
{{1 + x^2 }} - \frac{y}
{{1 + xy}}} \right\}\,dx} = \frac{{1}}
{2}\ln 2 + \frac{\pi }
{4}y - \ln (1 + y).$ This yields
$\displaystyle \lambda = \frac{{\ln 2}}
{2}\int_0^1 {\frac{1}
{{1 + y^2 }}\,dy} + \frac{\pi }
{4}\int_0^1 {\frac{y}
{{1 + y^2 }}\,dy} - \underbrace {\int_0^1 {\frac{{\ln (1 + y)}}
{{1 + y^2 }}\,dy} }_\lambda,$
and finally, after some simple calculations, we get $\displaystyle \lambda = \frac{1}
{2}\bigg\{ {\frac{\pi }
{8}\ln 2 + \frac{\pi }
{8}\ln 2} \bigg\} = \frac{\pi }
{8}\ln 2.$
The full answer to the proposed integral is $\displaystyle \frac{\pi }
{4}\ln 2 - \frac{\pi }
{8}\ln 2 = \frac{\pi }
{8}\ln 2,$ as required.
Anyway you should've wait TPH, since integral #6 is not solved yet.
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There're many other ways to tackle this integral, as a clever substitution, differentiation under the integral sign, etc.
$\displaystyle \int_0^1 \frac{\ln (x+1)}{x^2+1} dx$ let $\displaystyle \mu = \tan^{-1} x$.
Thus, $\displaystyle \int_0^{\pi/4} \ln (\tan \mu + 1) d\mu = \int_0^{\pi/4} \ln (\cos \mu + \sin \mu) d\mu -\int_0^{\pi/4} \ln (\cos \mu) d\mu$.
But, $\displaystyle \cos \mu + \sin \mu = \sqrt{2} \cos \left( \frac{\pi}{4} - \mu \right)$.
Thus, $\displaystyle \int_0^{\pi/4} \ln \left[ \sqrt{2} \cos \left( \frac{\pi}{4} - \mu \right) \right] d\mu - \int_0^{\pi/4} \ln (\cos \mu) d\mu$.
We get, $\displaystyle \frac{\pi}{8}\ln 2 + \int_0^{\pi/4} \ln \cos \left( \frac{\pi}{4} - \mu\right) \mu - \int_0^{\pi/4} \ln (\cos \mu) d\mu = \frac{\pi}{8}\ln 2$.
(Because last two integrals are the same).
Okay well, let's give another approach:
$\displaystyle \int_0^1 {\frac{{\ln (1 + x)}}
{{1 + x^2 }}\,dx} .$ Substitute $\displaystyle x = \frac{{1 - z}}
{{1 + z}},$ then the calculations give us $\displaystyle \int_0^1 {\frac{{\ln (1 + x)}}
{{1 + x^2 }}\,dx} = \int_0^1 {\frac{{\ln 2 - \ln (1 + z)}}
{{1 + z^2 }}\,dz} ,$ the rest follows.
This is an old one but hey?
Now this is just for an amazingly last ditch effort. but it shows the usefulness in last ditch efforts for series
Using polynomial division of $\displaystyle \frac{\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}}{\sum_{n=0}^{\infty}(-1)^nx^n}$ on the first 15 terms of that sequence and integrating termwise gives....let me see here
o! 0.2721953 [/tex]
$\displaystyle \frac{\pi}{8}\ln(2)\approx{.2721983}$
This should not be taken as a method to be used in such circumstances as this. But it does show that in a definite integral nightmare power series can be a ever-useful tool...but remember it rarely gives an exact answer...unless your good at seeing patterns
The main reason I did this, is that I am mostly ignorant of double integration and I wanted to find a non-power series method....but to no avail ...so this arose
...besides...I know I need to give everyone fuel to make fun of my power series use
Can you just explain to me few things ? And that's for this, or your previous series tutorial...
- are you basing all your maths on calculators or Maple ? I assume you didn't calculate this on your own...
- this is twice that you base your result onto the known result. What's the use of this ? Everyone can say "well, I'll approximate this function with another one, and I'll say it's more or less equal to the result others got"
- Where the heck is the beauty in it ?