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find the value of the integral $\int ^{\infty}_0\dfrac{e^{3x}}{(e^x+e^{5x})^2}dx$

$\int ^{\infty}_0\dfrac{e^{3x}}{(e^x+e^{5x})^2}dx=\int ^{\infty}_0\dfrac{e^{x}}{(1+e^{4x})^2}dx$

let $e^{x} = t \implies e^{x}dx=dt$

So

$$\int ^{\infty}_0\dfrac{e^{x}}{(1+e^{4x})^2}dx= \int ^{\infty}_0 \dfrac{dt}{(1+t^4)^2}$$

How to solve in terms of beta or gamma

UNAN
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Getheeka ponnada
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2 Answers2

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From this answer we get \begin{align*} \int_1^{\infty} \frac{\mathrm{d}x}{\left(1+x^4\right)^2} & =\frac x{4\left(x^4+1\right)}+\frac3{8\sqrt2} \text{arctan}\left(\frac{x^2-1}{\sqrt2x}\right) + \frac3{8\sqrt2} \text{arccoth}\left(\frac{x^2+1}{\sqrt2x}\right)\Bigg\vert_1^{\infty}\\ & = \boxed{\frac{3\pi}{16\sqrt{2}} - \frac{1}{8} - \frac{3}{8\sqrt{2}}\text{arccoth}\left(\sqrt{2}\right)} \end{align*}

Robert Lee
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  • Maple says $${\frac {3\pi}{16\sqrt2}}-\frac18-{\frac {3\ln \left( 3+2 ,\sqrt {2} \right) }{16\sqrt2}}$$ That is,$$\coth \left(\frac{\ln \left( 3+2,\sqrt {2} \right)}{2} \right)= \sqrt {2}$$ – GEdgar Mar 07 '22 at 16:38
  • Indeed! We can rewrite the result as you say exploiting $\frac{1}{2} \ln\Big\vert\frac{1+x}{1-x} \Big\vert = \text{arccoth}(x)$ for $|x|>1$, which can be nicely proven by integrating $\frac{1}{1-x^2}$ using hyperbolic trig-sub and using partial fractions. – Robert Lee Mar 07 '22 at 16:45
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From my post, $$ \int \frac{d x}{\left(1+x^{4}\right)^{2}}=\frac{x}{4\left(1+x^{4}\right)}+\frac{3}{16 \sqrt{2}}\left[2 \tan ^{-1}\left(\frac{x^{2}-1}{\sqrt{2} x}\right)+\ln \left|\frac{x^{2}+\sqrt{2} x+1}{x^{2}-\sqrt{2} x+1}\right|\right]+C $$ We can conclude that

$$ \int ^{\infty}_0\frac{e^{3x}}{(e^x+e^{5x})^2}dx \stackrel{t=e^x}{=} \int_1^{\infty} \frac{d t}{\left(1+t^{4}\right)^{2}}= \boxed{-\frac{1}{8}+\frac{3}{16 \sqrt{2}}\left[\pi- \ln (3+2\sqrt{2})\right]} $$

Lai
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