# $\int_ {0}^{\infty} \frac{(e^{3x}-e^x)dx}{x(e^x+1)(e^{3x}+1)}$ [duplicate]

$$\int_ {0}^{\infty} \frac{(e^{3x}-e^x) \ \mathrm dx}{x(e^x+1)(e^{3x}+1)}$$

I tried converting it to $$\int_ {0}^{\infty} \frac{\big((e^{3x}+1)-(e^x+1)\big) \ \mathrm dx}{x(e^x+1)(e^{3x}+1)}$$

integral-calculator.com says no antiderivative found.

I would like to see how it is solved by Feynman's Trick.

• WolframAlpha evaluates such to be $\approx0.549306$ Commented Aug 11, 2020 at 6:20
• @Akalanka OP already did that. That was also the only thing OP did, why are you repeating it? Commented Aug 11, 2020 at 6:31
• Your question is a duplicate of this question and this question. Although, Ninad's answer is amazing and I would prefer to keep it. Commented Aug 11, 2020 at 9:55
• @Axion004 as user Ninad Munsi suggested it may be solved by Feynman's trick ,so I want to see it as I don't have knowledge of multivariable calculus. Commented Aug 11, 2020 at 10:47
• As I probably should have mentioned, the Feynman trick solution is much harder. Just learn the basics of multiple integration (constant bounds), it will take you an hour. Also you can't say you don't know multivariable calculus, Feynman's trick is multivariable calculus, so you do know a little :) Commented Aug 11, 2020 at 11:37

You were on the right track. The integral separates out into

$$\int_0^\infty \frac{dx}{x}\left(\frac{1}{e^x+1}-\frac{1}{e^{3x}+1}\right) = \int_0^ \infty\int_1^3 \frac{e^{xy}}{(e^{xy}+1)^2}\:dy\:dx$$

Then we can swap the order of integration to get

$$\int_1^3 \frac{dy}{y}\frac{-1}{e^{xy}+1}\Biggr|_0^\infty = \int_1^3\frac{dy}{2y} = \frac{\log 3}{2}$$

• I don't understand what you did,I am in highschool. Commented Aug 11, 2020 at 6:34
• @user69608 it is not a highschool-level integral Commented Aug 11, 2020 at 6:35
• @user69608 you have to know some multivariable calculus for JEE, especially if you want a high math and physics score. This is not particularly complicated multivariable, either, because the bounds are all constants, nor is a multivariable change of variable necessary. Commented Aug 11, 2020 at 6:40
• ok thanks, though i suspect if there is a simpler method Commented Aug 11, 2020 at 6:41
• @user69608 I can guarantee you there is not, as I have taught multivariable calculus for a long time. This is a special case of a class intergrals known as Frullani Integrals. You can see that the proof immediately relies on multivariable calculus (particularly Fubini's theorem). In fact you even found that no elementary antiderivative exists, which is what almost all high school calculus relies on. Commented Aug 11, 2020 at 6:42

We can write $$I=\int_{0}^{\infty}\frac{\phi(x)-\phi(3x)}{x} dx, ~~~\phi(x)=\frac{1}{1+e^x}$$ Next we use the ideas of Frullani's Integration formula:

https://en.wikipedia.org/wiki/Frullani_integral

Since $$\int^{\infty}_0 \phi(x) dx$$ is convergent at $$\infty$$, then $$\int_{0}^{\infty} \frac{\phi(ax)-\phi(bx)}{x} dx=\phi_0 \ln \frac{b}{a},~~ \phi_0=\lim_{x\to 0} \phi(x)=\frac{1}{2}.$$ So $$I=\frac{\ln 3}{2}$$

I will present a simple solution using Feynman's method, as you asked for. Consider the following parameterized integral: $$I(a)=\int_{0}^{\infty} \frac{e^{ax}-e^x}{x\left(e^{ax}+1\right) \left(e^x+1\right)} \; dx$$ Now, differentiate both sides with respect to $$a$$ (factor out the terms independent of $$a$$ for simplicity): $$I'(a)=\int_{0}^{\infty} \frac{1}{x \left(e^x+1\right)} \cdot \frac{x e^{ax}\left(e^{ax}+1\right)-xe^{ax}\left(e^{ax}-e^x\right)}{{\left(e^{ax}+1\right)}^2} \; dx$$ Simplifying this yields an elegant integral: $$I'(a)=\int_{0}^{\infty} \frac{e^{ax}}{{\left(e^{ax}+1\right)}^2} \; dx$$ $$I'(a)=-\frac{1}{a\left(e^{ax}+1\right)} \bigg \rvert_{0}^{\infty}$$ $$I'(a)=\frac{1}{2a}$$ Integrate both sides with respect to $$a$$: $$I(a)=\frac{\ln{a}}{2}+C$$ If you plug in $$a=1$$ into the original, you get that $$I(1)=0$$. $$0=\frac{\ln{1}}{2}+C \implies C=0$$

Therefore, your integral is: $$\boxed{I=I(3)=\frac{\ln{3}}{2}}$$

• Why does $$\int_{0}^{\infty} \frac{e^{ax}}{{\left(e^{ax}+1\right)}^2} \; dx \implies\frac{1}{a\left(e^{ax}+1\right)} \bigg \rvert_{0}^{\infty}?$$ When I did the substitution $u=e^{ax}+1$, I found that the lower limit was $2$ instead of $0$ but I think that I screwed up. Is this step missing a negative sign: $$\int_{0}^{\infty} \frac{e^{ax}}{{\left(e^{ax}+1\right)}^2} \; dx \implies-\frac{1}{a\left(e^{ax}+1\right)} \bigg \rvert_{0}^{\infty}?$$ Commented Aug 11, 2020 at 16:21
• @Axion004 It is missing a negative sign but my work following it is using the correct result, I'll fix that. If $u=e^{ax}+1$ then $du=a e^{ax} dx$. Rewrite the integral as $$\int_2^{\infty} \frac{1}{a} \cdot \frac{du}{u^2} = -\frac{1}{au} \bigg \rvert_2^{\infty}$$ Here you still get $\frac{1}{2a}$ with the lower bound of $2$ and upper bound of $\infty$ after the substitution. In my answer I sort of did the substitution in my head and substituted $e^{ax}+1$ back for $u$ so that's why the lower bound is $0$ in my answer. it's the same thing.
– Ty.
Commented Aug 11, 2020 at 16:23
• @Axion004 I edited my previous comment explaining it. If something's unclear let me know.
– Ty.
Commented Aug 11, 2020 at 16:27
• That makes perfect sense. Thanks for answering these comments (+1). Commented Aug 11, 2020 at 16:38