Evaluating $\int_{0}^{t}\frac{e^{-x}\sin(x)}{x}dx$

Question

In my assignment of Laplace transforms, I have been given the task to evaluate the integral $$\int_{0}^{t}\frac{e^{-x}\sin(x)}{x}dx$$ The case when $t\to \infty$ can be easily tackled. But for any $t>0$, it seems rather absurd and tedious, and not at all related to Laplace transform. Here's what I have done.

$$\begin{aligned} I=\int_{0}^{t}\frac{e^{-x}\sin(x)}{x}dx&=\int_{0}^{t}\sum_{r=0}^{\infty}\frac{(-1)^{r}x^{r}}{r!}\frac{\sin(x)}{x}dx \\ &=\sum_{r=0}^{\infty}\frac{(-1)^{r}}{r!}\int_{0}^{t}x^{r-1}\sin(x)dx \end{aligned}$$

Now, using Integration by Parts, I arrived at the following result. $$\int_{0}^{t}x^{r-1}\sin(x)dx=\sum_{i=0}^{r-1}(-1)^{i}i!{r-1\choose i}t^{r-1-i}\sin\left(t+\frac{\pi}{2}(1+i)\right)$$ $$I=\sum_{r=0}^{\infty}\frac{(-1)^{r}}{r!}\sum_{i=0}^{r-1}(-1)^{i}i!{r-1\choose i}t^{r-1-i}\sin\left(t+\frac{\pi}{2}(1+i)\right)$$

Is there some way to solve this, and is the above evaluation correct. Thanks in advance.

Answer 1

While also not a Laplace transform, try using the real and imaginary part of $e^{ix}$ and splitting integrals:

$$\int_0^t \frac{\sin(x)}{xe^x}dx=\int_0^t\frac{\frac i2(e^{-ix}-e^{ix})}{xe^x}dx=\frac i2\left(\int_0^t \frac {e^{-(i+1)x}}xdx-\int_0^t \frac{e^{(i-1)x}}xdx\right)$$

Using the Exponential Integral $\text{Ei}(x)$:

$$\frac i2\left(\int_0^t \frac {e^{-(i+1)x}}xdx-\int_0^t \frac{e^{(i-1)x}}xdx\right)=\begin{cases}\frac i2(\text{Ei}(-(1+i)t)-\text{Ei}((i-1)t)-\frac{3\pi}4,t>0\\ \frac i2(\text{Ei}(-(1+i)t)-\text{Ei}((i-1)t)+\frac\pi4,t<0\end{cases}$$

where the integration is correct. The indefinite integral looks like a sine integral or a cosine integral function, but I could not get it into those forms. Also, the cases at the very end are due to $$\lim_{x\to0^\pm} \text{Ei}(-(1+i)x)-\text{Ei}((i-1)x)$$ having different results based on if $x<0$ or $x>0$ respectively. If you really want a series expansion, then just expand $e^x$ into one and integrate for $t>0$ to have positive terms:

$$\int \frac{e^{ax}}x dx=\int \frac 1x+\sum_{n=1}\frac{x^{n-1}}{n!}dx=C+\ln(x)+\sum_{n=1}^\infty \frac{x^n}{nn!}$$

Answer 2

$\textbf{Hint}$: The function

$$e^tf(t) = \int_0^t\frac{e^{t-x}\sin x}{x}dx$$

is a convolution. The Laplace transform of a convolution is the product of the Laplace transforms. Can you take it from here?

Answer 3

Let $\alpha_1(s)=\mathcal{L}\left(\frac{t+1}{(t+1)^2+1}\right)$ and $\alpha_2(s)=\mathcal{L}\left(\frac{1}{(t+1)^2+1}\right)$. Since $\frac{1}{x}=\int_0^{\infty}e^{-xy}\mathrm{d}y$ for any $x>0$ we can say for any $s>0$ that $$\begin{eqnarray*} \int_0^s \frac{e^{-x}\sin(x)}{x}\mathrm{d}x &=& \int_0^s\int_0^{\infty}e^{-x}\sin(x)e^{-xy}\mathrm{d}y\mathrm{d}x \\ &=& \int_0^{\infty}\int_0^se^{-x(y+1)}\sin(x)\mathrm{d}x\mathrm{d}y \\ &=& -\sin(s)\int_0^{\infty}\frac{(y+1)e^{-(y+1)s}}{(y+1)^2+1}\mathrm{d}y-\cos(s)\int_0^{\infty}\frac{e^{-(y+1)s}}{(y+1)^2+1}\mathrm{d}y+\int_0^{\infty}\frac{\mathrm{d}y}{(y+1)^2+1} \\ &=& -\sin\left(s\right)\mathcal{L}\left(u_{1}(t)\times\frac{t}{t^{2}+1}\right)-\cos(s)\mathcal{L}\left(u_1(t)\times \frac{1}{t^2+1}\right)+ \frac{\pi}{4} \\ &=&-\frac{\sin\left(s\right)}{e^s}\alpha_1(s)-\frac{\cos(s)}{e^s}\alpha_2(s)+ \frac{\pi}{4}\end{eqnarray*}$$

I feel this is a more useful description of this integral as it's now quite easy to establish the fact that $\int_0^{\infty}\frac{e^{-x}\sin(x)}{x}\mathrm{d}x=\frac{\pi}{4}$

Answer 4

You can simply evaluate the problem as suggested by @asgeige $$\mathcal{L}\left[\int_0^{t} \frac{e^{-x} sinx}{x}\right] dx$$

Let $g(t) = \frac{e^{-t} sint}{t}$ And from the fact that $$\mathcal{L} \left[\frac{f(t)}{t}\right] = \int_s^{\infty} F(y) dy$$ We can write $$G(s) = \int_s^{\infty} \frac{dy}{(y+1)^2 +1} dy$$ $$G(s) = \pi/2 - \tan^{-1}(s+1)$$ $$G(s) = cot^{-1}(s+1) $$ And from the property of Laplace transform of integral $$\mathcal{L}\left[\int_0^t f(\tau) d\tau \right] = \frac{F(s)}{s}$$

Combining the two results we get $$\mathcal{L}\left[\int_0^{t} \frac{e^{-x} sinx}{x}\right] dx = \frac{cot^{-1} (s+1)}{s}$$

Answer 5

The Laplace transform is defined as

$$F(s)=\mathcal{L}_x\left[f(x)\right](s)=\int\limits_0^\infty f(x)\, e^{-s\, x}\,dx\tag{1}$$

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Your integral

$$\int\limits_0^t \frac{e^{-x}\, \sin(x)}{x}\, dx\tag{2}$$

can be represented as the Laplace transform of

$$f(x)=\frac{\sin(x)}{x} (1-\theta(x-t))\tag{3}$$

evaluated at $s=1$ which leads to

$$F(1)=\int\limits_0^t \frac{e^{-x}\, \sin(x)}{x}\, dx=\mathcal{L}_x\left[\frac{\sin(x)}{x} (1-\theta(x-t))\right](1)=\int\limits_0^\infty \frac{\sin(x)}{x}\, (1-\theta (x-t))\, e^{-x} \, dx$$ $$=t \, _1F_4\left(\frac{1}{4};\frac{1}{2},\frac{3}{4},\frac{5}{4},\frac{5}{4};-\frac{t^4}{64}\right)+\frac{1}{18} t^2 \left(2 t \, _1F_4\left(\frac{3}{4};\frac{5}{4},\frac{3}{2},\frac{7}{4},\frac{7}{4};-\frac{t^4}{64}\right)-9 \, _1F_4\left(\frac{1}{2};\frac{3}{4},\frac{5}{4},\frac{3}{2},\frac{3}{2};-\frac{t^4}{64}\right)\right)\tag{4}$$

where $_1F_4$ is the HypergeometricPFQ function.

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The more general result is

$$F(s)=\mathcal{L}_x\left[\frac{\sin(x)}{x}\, (1-\theta(x-t))\right](s)=\int\limits_0^\infty \frac{\sin(x)}{x}\, (1-\theta (x-t))\, e^{-s\, x}\,dx$$ $$=-\frac{1}{2} i\, \text{Ei}(-((-i+s)\, t))+\frac{1}{2} i\, \text{Ei}(-((i+s)\, t))+\cot^{-1}(s)-\pi\tag{5}$$

which implies

$$F(1)=\frac{1}{2} i\, \text{Ei}((-1-i)\, t)-\frac{1}{2} i\, \text{Ei}((-1+i)\, t)-\frac{3 \pi }{4}\tag{6}$$

is equivalent to the result illustrated in formula (4) above.