how to solve the following definite integral?

Question

I am really confused about solving it.

$$\int_{-2}^2 \frac{x^2}{1+5^x} \, dx $$

Your first step (from $\int_{-2}^2$ to $2\int_0^2$) is invalid, since the integrand is not symmetric ($x^2$ is symmetric, but $1+5^x$ is not). — Arthur, Mar 22 '16 at 08:02
You don't have to delete your attempt, because it is sufficiently showing your effort. — choco_addicted, Mar 22 '16 at 08:09
I doubt there is an easy analytical solution. For a numerical one, though, try it out on wolfram alpha. — Vim, Mar 22 '16 at 08:12
Possible duplicate of Integrating $\int^2_{-2}\frac{x^2}{1+5^x}$ — Hans Lundmark, Mar 22 '16 at 09:59

score 2 · Accepted Answer · answered Mar 22 '16 at 08:13

2

HINT:

Use $\displaystyle I=\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$

$\displaystyle I+I=\int_a^bf(x)\ dx+\int_a^bf(a+b-x)\ dx=\int_a^b[f(x)+f(a+b-x)]\ dx$

answered Mar 22 '16 at 08:13

lab bhattacharjee

274,582

1

Bang on target!! Awesome Hint... – SchrodingersCat Mar 22 '16 at 08:16
@GautamChandna, Reminds me of http://math.stackexchange.com/questions/801278/how-to-solve-this-integral-int-11-fracx4ax1dx/801284#801284 and http://math.stackexchange.com/questions/741580/integral-int-pi-2-pi-2-frac12007x1-cdot-frac-sin2008x-si – lab bhattacharjee Mar 22 '16 at 08:22
@labbhattacharjee hmm so almost a duplicate, was looking for it for a long time, wasnt able to find it. thanks for your help again! – Gautam Chandna Mar 22 '16 at 08:28

KR136 · Answer 2 · 2016-03-22T08:53:39.057

0

There is a solution in special functions, achieved via use of substitutions.

$\int_{-2}^2 \frac{x^2}{1+5^x} \, dx$

Let $t=5^x$

$x= \log_5(t)$

$dx = \frac{dt}{\log_5(t) \ln(5)}$

$t(-2)=\frac{1}{25}; t(2)=25$

$\int_{-2}^2 \frac{x^2}{1+5^x} \, dx = $$\int_{\frac{1}{25}}^{25} \frac{(\log_5(t))^2}{1+t} \, \frac{dt}{\log_5(t) \ln(5)}$

$\int_{\frac{1}{25}}^{25} \frac{\log_5(t)}{1+t} \, \frac{dt}{ \ln(5)}$

$\frac{1}{ (\ln(5))^2} \int_{\frac{1}{25}}^{25} \frac{\ln(t)}{1+t} \, dt$

This last integral is special. It is closely related to the polylogarithm, and in particular, the dilogarithm.

By using the properties of these functions and the regular logarithm, I am sure you could reason out exactly what WolframAlpha tells us this is equal to:

$\frac{1}{ (\ln(5))^2} [Li_2(-t)+\log(t)\log(t+1)]^{25}_{\frac{1}{25}}$

$\frac{1}{ (\ln(5))^2} [[Li_2(-25)+\log(25)\log(26)]-[Li_2(-\frac{1}{25})+\log(\frac{1}{25})\log(\frac{26}{25})]]$ ≈1.493

edited Mar 22 '16 at 08:53

answered Mar 22 '16 at 08:19

KR136

1,810

Sorry, let me elaborate..one sec. – KR136 Mar 22 '16 at 08:20
Thats awesome! Learned some new stuff from your answer. Thanks alot! – Gautam Chandna Mar 22 '16 at 10:38
@GautamChandna No problem! Substitutions always surprise me- they so often are more useful than you think they'd be. – KR136 Mar 22 '16 at 13:15
Your solution gave me a new way to look at such problems! Thanks for it. – Gautam Chandna Mar 22 '16 at 18:49

how to solve the following definite integral?

2 Answers2