1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.