how to evaluate $\int \ln (\sqrt{e^x+\sin x}) dx$

Question

I'm newbie here, how to evaluate $$\int \ln (\sqrt{e^x+\sin x}) dx$$ I tried to use integration by parts but it didn't work.

Answer 1

There is no closed form for this integral so any technique you use won't work.

Edit: As you have now supplied a range you could evaluate the definite integral via numerial techniques such as the trapezoid rule or similar. However determining the indefinite integral is not the way to evaluate this problem.

High precision application of the trapezoid rules gives a definite integral (from 0 to 6) of: $8.86445077465226...$

Answer 2

This is not an answer but it is too long for a comment.

As already said in answers and comments, I do not think that there is any hope to get a closed form expression for the antiderivative.

However, assuming that you need to compute the value of the integral over a small range, you could expand the integrand as a Taylor series. This is not the easiest one to work, but, by the end, we could get $$\log \left(\sqrt{e^x+\sin (x)}\right)=x-\frac{3 x^2}{4}+\frac{5 x^3}{6}-\frac{25 x^4}{24}+\frac{17 x^5}{12}-\frac{361 x^6}{180}+\frac{184 x^7}{63}-\frac{1459 x^8}{336}+\frac{59497 x^9}{9072}-\frac{758203 x^{10}}{75600}+O\left(x^{11}\right)$$ So $$\int_0^a\log \left(\sqrt{e^x+\sin (x)}\right)\,dx=\frac{a^2}{2}-\frac{a^3}{4}+\frac{5 a^4}{24}-\frac{5 a^5}{24}+\frac{17 a^6}{72}-\frac{361 a^7}{1260}+\frac{23 a^8}{63}-\frac{1459 a^9}{3024}+\frac{59497 a^{10}}{90720}-\frac{758203 a^{11}}{831600}+O\left(a^{12}\right)$$

Let us try for $a=\frac{1}{10}$. The above formula gives $$\frac{594880155859783}{124740000000000000}\approx 0.00476896068510328$$ while an accurate numerical integration would lead to $$0.00476896068623222$$ The results become worse and worse (not to say more) when $a$ increases. For example, if $a=\frac 12$, the above formula gives $$\frac{261574073}{2554675200}\approx 0.102390344181523$$ while numerical integration would lead to $$0.102574149117685$$

For sure, you could add more terms to the expansion but this will not change the problem.

Edit

Looking at the plot of $\log \left(\sqrt{e^x+\sin (x)}\right)$ it differs from the plot of $\frac x 2$ mainly for $0<x<2\pi$. So, for any value of $a >2\pi$, we could approximate the value of the integral by $$I=\int_0^a\log \left(\sqrt{e^x+\sin (x)}\right)\,dx=\int_0^{2\pi} \Big(\log \left(\sqrt{e^x+\sin (x)}\right)-\frac x 2\Big)\,dx+\frac {a^2}4$$ that is to say, more or less,$$I\approx\frac {a^2}4+0.222885$$