This appears in my guide of definite integrals ;

I haven't seen complex analysis yet.. $$\int_1^e\sin{(\ln{(x)})}\mathrm{d}x\tag1$$
I used integration by parts and substitution 3 times in (1), but I get nowhere.. any help?

  • $\begingroup$ natural logarithm in base e, how to express that using mathjax? $\endgroup$ – Sebastian Fernandez Apr 14 at 22:26
  • $\begingroup$ Thanks for the edit. $\endgroup$ – Sebastian Fernandez Apr 14 at 22:28

If you use the substitution $u = \ln x$, then $du = \frac{1}{x} dx = e^{-u}dx$ and we get $$ I=\int_{1}^{e} \sin(\ln x) dx = \int_{0}^{1} e^{u}\sin u du. $$ This is a quite famous integral, and you can compute this by using the integration by parts twice. See here.


A slightly different way to see it: note that $x\in[1,e]$ in the integrand, hence you can multiply and divide by $x$, that is

$$\int_1^e\sin(\ln x)\, dx=\int_1^ex\frac{\sin(\ln x)}{x}\, dx\tag1$$

Then note that if we set $f(x):=-\cos x$ and $g(x):=\ln x$ then using the chain rule we have that

$$[-\cos(\ln x)]'=[(f\circ g)(x)]'=(f'\circ g)(x) g'(x)=\sin(\ln x)\frac1x\tag2$$

Thus we have that your integral is $-\int_1^e x[\cos(\ln x)]'\, dx$, so using integration by parts we get

$$-\int_1^ex[\cos(\ln x)]'\, dx=-x\cos(\ln x)\big|^e_1+\int_1^e\cos(\ln x)\, dx\tag3$$

Repeating the same method of above we find that

$$\int_1^e x\frac{\cos(\ln x)}x\, dx=x\sin(\ln x)\big|_1^e-\int_1^e \sin(\ln x)\, dx\tag4$$

and so

$$2\int_1^e\sin(\ln x)\, dx=x(\sin(\ln x)-\cos(\ln x))\big|_1^e=e(\sin 1-\cos 1)+1\tag5$$


Starting from @Seewoo Lee's answer $$I=\int_{1}^{e} \sin(\ln x) dx = \int_{0}^{1} e^{u}\sin( u)\, du$$ we can avoid integration by parts considering $$\int e^u \,e^{iu}\,du=\int e^{(1+i) u}\,du=\frac{1-i}2 e^{(1+i) u}$$ $$I=\int_0^1 e^u \,e^{iu}\,du=\frac{1-i}2 \left(e^{1+i}-1\right)$$ Expanding $$I==-\frac{1}{2}+\frac{1}{2} e \sin (1)+\frac{1}{2} e \cos (1)+i \left(\frac{1}{2}+\frac{1}{2} e \sin (1)-\frac{1}{2} e \cos (1)\right)$$ Now, consider the imaginary part.


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