6

$$I=12\int_{0}^{\infty}{x\over (e^{x^2}+1)(e^{x^2}+3)}dx=\ln(2)$$

$u=e^{x^2}\rightarrow du=2xe^{x^2}dx$

$x\rightarrow \infty,\, u=\infty$

$x\rightarrow 0,\, u=1$

$$6\int_{1}^{\infty}{2x\over (u+1)(u+3)}\cdot{du\over 2xe^{x^2}}=6\int_{1}^{\infty}{1\over u(u+1)(u+3)}du$$

$6=A(u+1)(u+3)+Bu(u+3)+Cu(u+1)$

$u=0\rightarrow A=2$

$u=-1\rightarrow B=-3$

$u=-3\rightarrow C=1$

$$I=\int_{1}^{\infty}{2\over u}-{3\over u+1}+{1\over u+3}dx$$

$$I=\left.\ln{u^2(u+3)\over (u+1)^3}\right|_{1}^{\infty}$$

$$I=\lim_{u\to \infty}\ln{u^2(u+3)\over (u+1)^3}-\ln{4\over 8}$$

$$I=\ln(2)$$

Another quick method of approach at tackling I?

snulty
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gymbvghjkgkjkhgfkl
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    This seems to me the most straightforward approach to the problem. Someone else might have a very clever insight, but this is probably the most reliable method. – Joel Jun 02 '16 at 14:54
  • Hint: Integrate $$ \frac{\log(1-z)}{z(z+1)(z+3)}$$ around a keyhole in the complex plane, if u don't like partial fractions – tired Jun 02 '16 at 15:06
  • You could generalize and get in the same manner $$\int_{0}^{\infty}{x\over (e^{x^2}+a)(e^{x^2}+b)}dx=\frac{a \log (b+1)-b \log (a+1)}{2 a b (a-b)}$$ assuming $a\neq b$, $a\geq 0$, $b\geq 0$. – Claude Leibovici Jun 02 '16 at 15:07
  • Please improve the quality of the post by adding exposition, in natural language. – Carl Mummert Jun 02 '16 at 17:47
  • Several of your other recent questions have also been of relatively low expositional quality. Good questons include context, as described here: http://meta.math.stackexchange.com/questions/9959/how-to-ask-a-good-question#9960 . Each question should stand alone without the title, and should include a clear statement of what is being asked, – Carl Mummert Jun 02 '16 at 17:54

2 Answers2

2

Following Claude Leibovici's comment, assuming $\text{Re}(a),\text{Re}(b)>-1$,

$$\begin{eqnarray*}J(a,b)=\int_{0}^{+\infty}\frac{2x\,dx}{(e^{x^2}+a)(e^{x^2}+b)}&=&\int_{0}^{+\infty}\frac{du}{(e^{u}+a)(e^{u}+b)}\\&=&\frac{1}{b-a}\int_{0}^{+\infty}\left(\frac{1}{e^u+a}-\frac{1}{e^u+b}\right)\,du\\&=&\frac{1}{b-a}\int_{1}^{+\infty}\frac{dx}{x}\left(\frac{1}{x+a}-\frac{1}{x+b}\right)\\&=&\frac{1}{b-a}\int_{0}^{1}\left(\frac{1}{1+ax}-\frac{1}{1+bx}\right)\,dx\\&=&\color{red}{\frac{b\log(a+1)-a\log(b+1)}{ab(b-a)}}\end{eqnarray*}$$ and by taking the limit as $b\to a$, $$ J(a,a) = \color{red}{\frac{(a+1)\log(a+1)-a}{a^2(a+1)}}.$$

Jack D'Aurizio
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1

$$I=12\int_{0}^{\infty}{x\over (e^{x^2}+1)(e^{x^2}+3)}dx=6\int_{0}^{\infty}{2xe^{x^2}\over e^{x^2}[(e^{x^2}+2)^2-1]}dx=6\int_{3}^{\infty}\frac{du}{(u-2)(u^2-1)}$$ $$I=2\int_{3}^{\infty}\left(\frac{1}{u-2}-\frac{u+2}{u^2-1}\right)du$$

Behrouz Maleki
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