$\int_{0}^{1} \sin x \cdot \ln x dx$ I would like to evaluate the following integral:
$$\int_{0}^{1} \sin x \cdot \ln x dx$$
I had a pretty decent idea to do it, and that is by turning $\sin x$ an $\ln x$ into their respective power series, turning it into a double summation, and integrating. This yields the following:
$$-\int_{0}^{1}\sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{x^{2n+1}(-1)^n(1-x)^{m+1}}{(m+1)(2n+1)!} dx$$
Putting the integral inside the double summation and making use the beta function, we get:
$$ =-\sum_{n=0}^{\infty} \sum_{m=0}^{\infty}\frac{(-1)^n \beta(2n+1,m+1)}{(m+1)(2n+1)!}$$
A slight reindexing gives:
$$\sum_{n \space \text{odd}} \sum_{m \space\geq 1}\frac{\beta(n,m)}{n!m}$$
Notice the $-1$ in the front goes away by cancelling with the $-1$ from letting $n$ run through the odds.
Is this a valid strategy? If there is a closed form for this integral, I think this is a pretty cool summation identity. Can someone help out?
 A: Multiple comments have already pointed to a special function $\operatorname{Cin}(x):=\int_0^x\frac{1-\cos t}{t}dt$. Integration by parts with $u:=\ln x,\,v:=1-\cos x$ gives $\int_0^1\sin x\ln xdx=[(1-\cos x)\ln x]_0^1-\operatorname{Cin}(1)$. The surface term is $-\lim_{x\to0^+}(1-\cos x)\ln x=-\tfrac12\lim_{x\to0^+}x^2\ln x=0$, so the original integral is $-\operatorname{Cin}(1)$.
A: Integrating by parts
$$\int \log x \sin x \, dx=-\cos x \log x-\int\left(- \frac{\cos x}{x}\right)\,dx=$$
$$=-\cos x \log x+\text{Ci}(x)+C$$
The requested improper integral is
$$\int_0^1 \log x \sin x \, dx=\text{Ci}(1)-\underset{x\to 0}{\text{lim}}(\text{Ci}(x)-\log x \cos x)=^*\text{Ci}(1)-\gamma$$
where $\gamma$ is the Euler-Mascheroni constant.
$^*$ MacLaurin expansions are
$\text{Ci}(x)=\log x+\gamma +O\left(x^2\right)$
$\log x\cos x=\log x+O\left(x^2\right)$
A: Slightly different approach
Consider a parameter defined integral:
$$I(a)=\int_{0}^1x^a\sin x\;dx$$
We'll get back to the original integral after differentiating this wrt $a$ and setting $a=0$
Next, use the Taylor series of $\sin x$:
$$I(a)=\int_{0}^1\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k+1+a}}{(2k+1)!}\;dx$$
$$=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!(2k+2+a)}$$
Now, differentiate wrt $a$ and set $a=0$:
$$\int_{0}^1\ln x\sin x\;dx=-\frac{1}{4}\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!(k+1)^2}$$
This infinite series converges very fast.
If we take only the first term of the series, $-\frac{1}{4}$, then there is an error only about $0.01$.
The sum of the first $2$ terms of the series is $-\frac{23}{96}$ and our deviation from the exact value is only about $0.00023$
