111
$\begingroup$

Compute the following integral \begin{equation} \int_0^{\Large\frac{\pi}{4}}\left[\frac{(1-x^2)\ln(1+x^2)+(1+x^2)-(1-x^2)\ln(1-x^2)}{(1-x^4)(1+x^2)}\right] x\, \exp\left[\frac{x^2-1}{x^2+1}\right]\, dx \end{equation}

I was given two integral questions by my teacher. I can answer this one although it took a lot of time to compute it. I want to share this problem to the other users here and I would love to see how Mathematics SE users compute this monster. Thank you.

$\endgroup$
76
$\begingroup$

Rewrite \begin{align} &\int\left[\frac{(1-x^2)\ln(1+x^2)+(1+x^2)-(1-x^2)\ln(1-x^2)}{(1-x^4)(1+x^2)}\right] x\ \exp\left[\frac{x^2-1}{x^2+1}\right]\ dx\\ &=-\int\left[\frac{(1-x^2)\ln\left(\dfrac{1-x^2}{1+x^2}\right)-(1+x^2)}{(1-x^2)(1+x^2)(1+x^2)}\right] x\ \exp\left[-\frac{1-x^2}{1+x^2}\right]\ dx\\ &=-\frac14\int\left[\frac{(1-x^2)\ln\left(\dfrac{1-x^2}{1+x^2}\right)-(1+x^2)}{(1-x^2)}\right] \frac{2x}{1+x^2}\, \exp\left[-\frac{1-x^2}{1+x^2}\right]\ \frac{2\ dx}{1+x^2}\\ &=-\frac14\int\left[\ln\left(\frac{1-x^2}{1+x^2}\right)-\frac{1+x^2}{1-x^2}\right] \frac{2x}{1+x^2}\, \exp\left[-\frac{1-x^2}{1+x^2}\right]\ \frac{2\ dx}{1+x^2}.\tag1 \end{align} Now, consider Weierstrass substitution: $$ x=\tan\frac{t}{2}\;,\;\sin t=\frac{2x}{1+x^2}\;,\;\cos t=\frac{1-x^2}{1+x^2}\;,\;\text{ and }\;dt=\frac{2\ dx}{1+x^2}. $$ The integral in $(1)$ turns out to be $$ -\frac14\int\left[\ln\left(\cos t\right)-\frac{1}{\cos t}\right] \sin t\, \exp\left[-\cos t\right]\ dt.\tag2 $$ Let $y=\cos t\;\Rightarrow\;dy=-\sin t\ dt$, then $(2)$ becomes $$ \frac14\int\left[\ln y-\frac{1}{y}\right] e^{-y}\ dy=\frac14\left[\int e^{-y}\ln y\ dy-\int\frac{e^{-y}}{y}\ dy\right].\tag3 $$ The second integral in the RHS $(3)$ can be evaluated by using IBP. Taking $u=e^{-y}\;\Rightarrow\;du=-e^{-y}\ dy$ and $dv=\dfrac1y\ dy\;\Rightarrow\;v=\ln y$, then $$ \int\frac{e^{-y}}{y}\ dy=e^{-y}\ln y+\int e^{-y}\ln y\ dy.\tag4 $$ Substituting $(4)$ to $(3)$, we obtain $$ \frac14\left[\int e^{-y}\ln y\ dy-e^{-y}\ln y-\int e^{-y}\ln y\ dy\right]=-\frac14e^{-y}\ln y+C. $$ Thus \begin{align} &\int\left[\frac{(1-x^2)\ln(1+x^2)+(1+x^2)-(1-x^2)\ln(1-x^2)}{(1-x^4)(1+x^2)}\right] x\ \exp\left[\frac{x^2-1}{x^2+1}\right]\ dx\\ &=\color{blue}{-\frac14\exp\left[-\frac{1-x^2}{1+x^2}\right]\ln \left|\frac{1-x^2}{1+x^2}\right|+C} \end{align} and \begin{align} &\int_0^{\Large\frac\pi4}\left[\frac{(1-x^2)\ln(1+x^2)+(1+x^2)-(1-x^2)\ln(1-x^2)}{(1-x^4)(1+x^2)}\right] x\ \exp\left[\frac{x^2-1}{x^2+1}\right]\ dx\\ &=\color{blue}{-\frac14\exp\left[\frac{\pi^2-16}{\pi^2+16}\right]\ln \left|\frac{16-\pi^2}{16+\pi^2}\right|}. \end{align}

| cite | improve this answer | |
$\endgroup$
  • 13
    $\begingroup$ I choose your answer since it's very detail and it's also nice-formatted. Thanks! (>‿◠)✌ $\endgroup$ – Anastasiya-Romanova 秀 Jun 1 '14 at 14:13
  • 8
    $\begingroup$ For prettiness factor why not write $\exp\left[\frac{\pi^2-16}{\pi^2+16}\right]$? $\endgroup$ – Ali Caglayan Dec 17 '14 at 14:52
76
$\begingroup$

From the numerator, collect the logarithmic terms first.

$$\displaystyle \int_0^{\pi/4} x\frac{(1+x^2)+(x^2-1)\ln\left(\frac{1-x^2}{1+x^2}\right)}{(1-x^4)(1+x^2)}\exp\left(\frac{x^2-1}{x^2+1}\right)\,dx$$

Rewrite $(1-x^4)=(1-x^2)(1+x^2)$ and divide the numerator by $(1+x^2)$.

$$\displaystyle \int_0^{\pi/4} \frac{x}{(1-x^2)(1+x^2)}\left(1+\frac{x^2-1}{x^2+1}\ln\left(\frac{1-x^2}{1+x^2}\right)\right)\exp\left(\frac{x^2-1}{x^2+1}\right)\,dx $$

Use the substitution $\displaystyle \frac{x^2-1}{x^2+1}=t \Rightarrow \frac{4x}{(1+x^2)^2}\,dx=dt$ to get:

$$\displaystyle \frac{1}{4}\int_{a}^{-1}\frac{e^t}{t}\left(1+t\ln(-t)\right)\,dt= \frac{1}{4}\int_{a}^{-1}e^t\left(\frac{1}{t}+\ln(-t)\right)\,dt$$

where $\displaystyle a=\frac{\pi^2/16-1}{\pi^2/16+1}$

Since $\displaystyle \int e^x(f'(x)+f(x))\,dx=e^xf(x)+C $, the above definite integral is:

$$\displaystyle \frac{1}{4}\left(e^t \ln(-t) \right|_{a}^{-1}=-\frac{1}{4}e^a\ln(-a) \approx \boxed{0.284007} $$

| cite | improve this answer | |
$\endgroup$
  • 31
    $\begingroup$ That was fast... $\endgroup$ – IAmNoOne May 31 '14 at 10:33
  • 6
    $\begingroup$ The FLASH!! LOL $\endgroup$ – Anastasiya-Romanova 秀 May 31 '14 at 10:34
  • 7
    $\begingroup$ @Nameless: Yeah, the reason being that V-Moy recently posted this problem on a problem solving website and I posted the very same solution there so all I had to was just copy-paste. ;) $\endgroup$ – Pranav Arora May 31 '14 at 10:34
  • $\begingroup$ @Nameless Please don't be so surprise. I've posted this question on Brilliant and Pranav has already answer it there $\ddot\smile$ $\endgroup$ – Anastasiya-Romanova 秀 May 31 '14 at 10:35
  • 1
    $\begingroup$ @Nameless Here is the link to the problem. $\endgroup$ – Anastasiya-Romanova 秀 May 31 '14 at 10:36
24
$\begingroup$

The Monster integral asked by Anastasiya Ramanova two years ago is of the form $$\color{blue}{ \begin{equation*} \int_{0}^{\frac{\pi }{4}}\left[ \frac{(1-x^{2})\ln (1+x^{2})+(1+x^{2})-(1-x^{2})\ln (1-x^{2})}{(1-x^{4})(1+x^{2})}\right] xe^{\left( \frac{x^{2}-1}{x^{2}+1}\right) }dx=\int_{0}^{\frac{\pi }{4} }h(x)e^{g(x)}dx. \end{equation*} }$$ There are already two answers each one uses a substitution. I would like to share mine since it is systematic and do not use any substitution.

The presence of the exponential function in the integrand function recalls the well-known formula $$\color{red}{ \begin{equation*} \int \left( f^{\prime }(x)+g^{\prime }(x)f(x)\right) e^{g(x)}dx=f(x)e^{g(x)}+C \end{equation*} }$$ with $\color{blue}{g(x)=\frac{x^{2}-1}{x^{2}+1}.}$

Its proof maybe found at

Proof for formula $\int e^{g(x)}[f'(x) + g'(x)f(x)] dx = f(x) e^{g(x)}+C$

So we are done if we find a function $\color{blue}{f(x)}$ such that

$$\color{blue}{ \begin{equation*} h(x)=f^{\prime }(x)+g^{\prime }(x)f(x). \end{equation*}}$$

In what follows, I will show step by step that $\color{blue}{f(x)=\frac{1}{4}\ln \left( \frac{1+x^{2}}{1-x^{2}}\right), }$ and therefore

$$\color{blue}{ \begin{equation*} \int\left[ \frac{(1-x^{2})\ln (1+x^{2})+(1+x^{2})-(1-x^{2})\ln (1-x^{2})}{(1-x^{4})(1+x^{2})}\right] xe^{\left( \frac{x^{2}-1}{x^{2}+1}\right) }dx=\frac{1}{4}\ln \left( \frac{ 1+x^{2}}{1-x^{2}}\right) e^{\left( \frac{x^{2}-1}{x^{2}+1}\right) }+C. \end{equation*} }$$ $\color{red}{\bf Problem:}$ We want to write $$\color{blue}{ \begin{equation*} \left[ \frac{(1-x^{2})\ln (1+x^{2})+(1+x^{2})-(1-x^{2})\ln (1-x^{2})}{ (1-x^{4})(1+x^{2})}\right] x \end{equation*}}$$ as $$\color{blue}{ \begin{equation*} f^{\prime }(x)+g^{\prime }(x)f(x) \end{equation*}}$$ where $$\color{blue}{ \begin{equation*} g(x)=\left( \frac{x^{2}-1}{x^{2}+1}\right) ,\ \ \ \ \ \ \ g^{\prime }(x)= \frac{4x}{(1+x^{2})^{2}} \end{equation*}}$$ and $\color{blue}{f(x)}$ is to be determined.

The unique thing which is given by the statement is $\color{blue}{g(x)=\left( \frac{x^{2}-1}{ x^{2}+1}\right) .}$ So, the first thing we start with is to look for $\color{red}{g^{\prime }(x)}$ inside what would be $\color{blue}{f^{\prime }(x)+\color{red}{g^{\prime }(x)}f(x),}$ that is, inside $$\color{blue}{ \begin{equation*} \left[ \frac{(1-x^{2})\ln (1+x^{2})+(1+x^{2})-(1-x^{2})\ln (1-x^{2})}{ (1-x^{4})(1+x^{2})}\right] x. \end{equation*} }$$ Let us break down this expression into three fractions as follows $$\color{blue}{ \begin{equation*} \left[ \frac{(1-x^{2})\ln (1+x^{2})}{(1-x^{4})(1+x^{2})}+\frac{(1+x^{2})}{ (1-x^{4})(1+x^{2})}-\frac{(1-x^{2})\ln (1-x^{2})}{(1-x^{4})(1+x^{2})}\right]x. \end{equation*} }$$ Next cancelling $$\color{blue}{ \begin{equation*} \left[ \frac{\ln (1+x^{2})}{(1+x^{2})(1+x^{2})}+\frac{1}{(1-x^{4})}-\frac{ \ln (1-x^{2})}{(1+x^{2})(1+x^{2})}\right] x. \end{equation*} }$$ The first and third fractions have the same denominator and moreover it is that of $\color{blue}{ g^{\prime }(x),}$ so we bring them together! $$\color{blue}{ \begin{equation*} \left[ \frac{1}{(1-x^{4})}+\frac{\ln (1+x^{2})-\ln (1-x^{2})}{ (1+x^{2})(1+x^{2})}\right] x. \end{equation*} }$$ Since $\color{blue}{\ln (1+x^{2})-\ln (1-x^{2})=\ln \left( \frac{1+x^{2}}{1-x^{2}}\right) ,}$ we can make $\color{blue}{g^{\prime }(x)=\color{red}{\frac{4x}{(1+x^{2})^{2}}}}$ to appear as follows $$\color{blue}{ \begin{equation*} \left[ \frac{x}{(1-x^{4})}+\color{red}{\frac{4x}{(1+x^{2})^{2}}}\frac{1}{4}\ln \left( \frac{1+x^{2}}{1-x^{2}}\right) \right] . \end{equation*} }$$ So, define $$\color{blue}{ \begin{equation*} f(x)=\frac{1}{4}\ln \left( \frac{1+x^{2}}{1-x^{2}}\right) \end{equation*} }$$

and compute $\color{blue}{f^{\prime }(x)}$ to find that $$\color{blue}{ \begin{equation*} f^{\prime }(x)=\frac{x}{1-x^{4}}. \end{equation*}}$$

It follows that the monster integral is nothing but $$\color{blue}{ \begin{equation*} \int_{0}^{\frac{\pi }{4}}\left[ f^{\prime }(x)+g^{\prime }(x)f(x)\right] e^{g(x)}dx=\left. f(x)e^{g(x)}\right] _{0}^{\frac{\pi }{4}}=f(\frac{\pi }{4} )e^{g(\frac{\pi }{4})}-f(0)e^{g(0)}. \end{equation*} }$$

Since $$\color{blue}{ \begin{equation*} f\left( \frac{\pi }{4}\right) =\frac{1}{4}\ln \left( \frac{1+\left( \frac{ \pi }{4}\right) ^{2}}{1-\left( \frac{\pi }{4}\right) ^{2}}\right) =\frac{1}{4 }\ln \left( \frac{16+\pi ^{2}}{16-\pi ^{2}}\right) ,\ \ \ \ \ \ \ \color{black}{\text{and}}\ \ \ \ \ \ \ \ \ \ f(0)=\frac{1}{4}\ln \left( \frac{1+0^{2}}{1-0^{2}}\right) =0 \end{equation*} } $$

$$\color{blue}{ \begin{equation*} g\left( \frac{\pi }{4}\right) =\left( \frac{\left( \frac{\pi }{4}\right) ^{2}-1}{\left( \frac{\pi }{4}\right) ^{2}+1}\right) =\frac{\pi ^{2}-16}{\pi ^{2}+16},\ \ \ \ \color{black}{\text{and}}\ \ \ \ \ \ \ \ \ \ \ \ g(0)=-1, \end{equation*} }$$ then $$\color{blue}{ \begin{equation*} \int_{0}^{\frac{\pi }{4}}\left[ \frac{(1-x^{2})\ln (1+x^{2})+(1+x^{2})-(1-x^{2})\ln (1-x^{2})}{(1-x^{4})(1+x^{2})}\right] xe^{\left( \frac{x^{2}-1}{x^{2}+1}\right) }dx = \frac{1}{4}\ln \left( \frac{16+\pi ^{2}}{16-\pi ^{2}}\right) e^{\left( \frac{\pi ^{2}-16}{\pi ^{2}+16} \right) }.\ \ \ \color{red} \blacksquare \end{equation*}}$$

| cite | improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ Nitpick: there's no point in blue font if everything is in blue font. Just makes it harder on my eyes personally. $\endgroup$ – Simply Beautiful Art Oct 11 '18 at 13:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.