So far on this site, I have seen many different types of substitutions being used for integration, such as the trigonometric functions, the exponential function $\exp(x)$, even $\tan(\frac{x}{2})$ and Euler's substitution. But when is it a good idea to consider a substitution of $\ln x$ (other than some obvious things like $\int\frac{\ln x}{x}dx$)? I found out that if I substitute $\ln x$ into an integral of the form $$\int_0^\infty f(x)dx$$then we get $$\int_{-\infty}^\infty\frac{f(\ln x)}{e^x}dx$$And then we could use the semicircle contour (with a singularity at $0$) to solve the integral using the Residue theorem. Any ideas?
Asked
Active
Viewed 56 times
0
-
1How about the integral of $e^{-x}$ on $[0,+\infty[$ ? – Hamdiken Feb 15 '23 at 18:31
-
@Hamdiken so this converts the integral into one between bounds $0$ and $1$. Am I right? – Kamal Saleh Feb 15 '23 at 18:32
-
And what function are you integrating ? – Hamdiken Feb 15 '23 at 18:39
-
Would this help? – Accelerator Feb 15 '23 at 18:47
-
@Accelerator Where is the substitution? Sorry, I am a bit blind. – Kamal Saleh Feb 15 '23 at 19:03
-
@Hamdiken Oh I got confused, I thought you were substituting $e^{-x}$. Thanks! – Kamal Saleh Feb 15 '23 at 19:03
-
1In my answer where I have $\int_{0}^{\infty}\frac{\left(e^{x}-1\right)e^{x}}{x\left(e^{2x}+1\right)e^{x}}dx = \int_{1}^{\infty}\left(\frac{x-1}{x\left(1+x^{2}\right)\ln\left(x\right)}\right)dx$, I used $u=e^x$. @KamalSaleh – Accelerator Feb 15 '23 at 19:12